It is worth considering the simplest non-abelian example more closely. The integer Heisenberg group $H_{\mathbb Z}$ is the simplest non-trivial example of a nilpotent group. It is generated by two elements $u,v$ with the relations that the commutator $w = uvu^{-1}v^{-1}$ commutes with $u$ and $v$. Explicitly, it is the group of $3\times 3$ upper-triangular matrices with integer entries and diagonal entries 1: take $u$ and $v$ to be the matrices $$u = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{array} \right),\ v = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\0 & 0 & 1 \end{array} \right),$$ so that $$w = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{array} \right).$$ Any element of $H_{\mathbb Z}$ can be written uniquely in the form $u^k v^l w^m$ for some integers $k,l,m$. The group $H_{\mathbb Z}$ sits inside the more usual 3-dimensional Heisenberg group $H$ consisting of the strictly upper-triangular matrices with real entries $$\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\0 & 0 & 1 \end{array} \right).$$ By analogy with the case of $\mathbb Z^2$ in the plane, you might expect the group $H_{\mathbb Z}$ to converge to $H$ under re-scaling. As a manifold, the group $H$ is just $\mathbb R^3$, so you would predict that $H_{\mathbb Z}$ has cubic polynomial growth just like the abelian group $\mathbb{Z}^3$. But actually it has \emph{quartic} growth. This is easy to see: because $vu = uvw$ we have $v^m u^n = u^n v^m w^{mn}$ for any $m,n$, and so any of the $n^4$ elements $u^a v^b w^c$ with $0 \leq a \leq n$, $0 \leq b \leq n$, and $0 \leq c \leq n^2$ can be obtained as the product of a string of at most $n$ copies of $u$ and at most $n$ copies of $v$ arranged in a suitable order. Why do we get quartic growth from such a 3-dimensional group? The answer takes us to a fascinating piece of geometry. Although $H$ is identified with $\mathbb R^3$, we find that if we want left-multiplications in the group to be isometries we must warp the usual metric of $\mathbb R^3$ somewhat. In fact the length of a path must be defined as the integral of $$\{dx^2 \ + \ dy^2 \ + \ (dz - xdy)^2 \}^{1/2},$$ which means that each plane perpendicular to the $x$-axis has been sheared in the $y$-direction by an amount proportional to $x$. As we re-scale the word-metric on $H_{\mathbb Z}$ we must re-scale the metric on $\mathbb R ^3$ too, multiplying $x$ and $y$ by $\lambda = 1/r_i$, and $z$ by $\lambda^2$. In the limit, this means that the only paths of finite length are those whose direction at each point lies in the plane given by $dz - xdy = 0$. Let us call them the \emph{allowable} paths. They give us a new metric on $\mathbb R^3$ in which the distance from one point to another is the length of the shortest allowable path joining them. This metric defines the usual topology on $\mathbb R^3$, but, unlike the taxi-cab metric, it is far from equivalent --- even locally --- to the usual metric. Nevertheless, it is a metric which those of us who have tried to park a car know all too well. To move a car just a little bit sideways we must take it along a disproportionately long path. This is because the position of a car sitting on an expanse of tarmac is described by three coordinates $(\xi,\eta,\theta)$, where $(\xi,\eta)$ is the position of the mid-point of the front wheels, and $\theta$ is the angle in which the axis of symmetry of the car is pointing. When we want to move the car we can move $\xi$ and $\eta$ any way we like, within reason, but the change in the third coordinate is constrained by the differential relation $d\theta = \cos \theta \ d\eta \ - \ \sin \theta \ d\xi,$ assuming the car is of unit length. (A simple change of coordinates puts this in the form $dz = ydx$ which we found for the group $H$.) What is remarkable about this new metric is that on one hand it defines the usual topology of $\mathbb R^3$, but on the other hand it defines a metric space of \emph{Hausdorff} dimension \emph{four}. Hausdorff dimension is a concept defined only for metric spaces. To say it is four means, essentially, that the number of balls of radius $r/n$ needed to cover a ball of radius $r$ grows like $n^4$ as $n \to \infty$. In the new metric on $\mathbb R^3$ this is the case because a small $\varepsilon$-ball for the new metric looks (in the usual coordinates) like a very flat ellipsoid with axes $(\varepsilon, \varepsilon, \varepsilon^2)$. This explains the quartic growth rate of the group $H_{\mathbb Z}$. More generally, for any metric space the topological dimension is bounded above by the Hausdorff dimension, and so the finite-dimensionality of the limit space $Y$ follows from the polynomial growth of the group.
It is worth considering the simplest non-abelian example more closely. The integer Heisenberg group $H_{\mathbb Z}$ is the simplest non-trivial example of a nilpotent group. It is generated by two elements $u,v$ with the relations that the commutator $w = uvu^{-1}v^{-1}$ commutes with $u$ and $v$. Explicitly, it is the group of $3\times 3$ upper-triangular matrices with integer entries and diagonal entries 1: take $u$ and $v$ to be the matrices $$u = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{array} \right),\ v = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\0 & 0 & 1 \end{array} \right),$$ so that $$w = \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{array} \right).$$ Any element of $H_{\mathbb Z}$ can be written uniquely in the form $u^k v^l w^m$ for some integers $k,l,m$. The group $H_{\mathbb Z}$ sits inside the more usual 3-dimensional Heisenberg group $H$ consisting of the strictly upper-triangular matrices with real entries $$\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\0 & 0 & 1 \end{array} \right).$$ By analogy with the case of $\mathbb Z^2$ in the plane, you might expect the group $H_{\mathbb Z}$ to converge to $H$ under re-scaling. As a manifold, the group $H$ is just $\mathbb R^3$, so you would predict that $H_{\mathbb Z}$ has cubic polynomial growth just like the abelian group $\mathbb{Z}^3$. But actually it has \emph{quartic} growth. This is easy to see: because $vu = uvw$ we have $v^m u^n = u^n v^m w^{mn}$ for any $m,n$, and so any of the $n^4$ elements $u^a v^b w^c$ with $0 \leq a \leq n$, $0 \leq b \leq n$, and $0 \leq c \leq n^2$ can be obtained as the product of a string of at most $n$ copies of $u$ and at most $n$ copies of $v$ arranged in a suitable order. Why do we get quartic growth from such a 3-dimensional group? The answer takes us to a fascinating piece of geometry. Although $H$ is identified with $\mathbb R^3$, we find that if we want left-multiplications in the group to be isometries we must warp the usual metric of $\mathbb R^3$ somewhat. In fact the length of a path must be defined as the integral of $$\{dx^2 \ + \ dy^2 \ + \ (dz - xdy)^2 \}^{1/2},$$ which means that each plane perpendicular to the $x$-axis has been sheared in the $y$-direction by an amount proportional to $x$. As we re-scale the word-metric on $H_{\mathbb Z}$ we must re-scale the metric on $\mathbb R ^3$ too, multiplying $x$ and $y$ by $\lambda = 1/r_i$, and $z$ by $\lambda^2$. In the limit, this means that the only paths of finite length are those whose direction at each point lies in the plane given by $dz - xdy = 0$. Let us call them the \emph{allowable} paths. They give us a new metric on $\mathbb R^3$ in which the distance from one point to another is the length of the shortest allowable path joining them. This metric defines the usual topology on $\mathbb R^3$, but, unlike the taxi-cab metric, it is far from equivalent --- even locally --- to the usual metric. Nevertheless, it is a metric which those of us who have tried to park a car know all too well. To move a car just a little bit sideways we must take it along a disproportionately long path. This is because the position of a car sitting on an expanse of tarmac is described by three coordinates $(\xi,\eta,\theta)$, where $(\xi,\eta)$ is the position of the mid-point of the front wheels, and $\theta$ is the angle in which the axis of symmetry of the car is pointing. When we want to move the car we can move $\xi$ and $\eta$ any way we like, within reason, but the change in the third coordinate is constrained by the differential relation $d\theta = \cos \theta \ d\eta \ - \ \sin \theta \ d\xi,$ assuming the car is of unit length. (A simple change of coordinates puts this in the form $dz = ydx$ which we found for the group $H$.) What is remarkable about this new metric is that on one hand it defines the usual topology of $\mathbb R^3$, but on the other hand it defines a metric space of \emph{Hausdorff} dimension \emph{four}. Hausdorff dimension is a concept defined only for metric spaces. To say it is four means, essentially, that the number of balls of radius $r/n$ needed to cover a ball of radius $r$ grows like $n^4$ as $n \to \infty$. In the new metric on $\mathbb R^3$ this is the case because a small $\varepsilon$-ball for the new metric looks (in the usual coordinates) like a very flat ellipsoid with axes $(\varepsilon, \varepsilon, \varepsilon^2)$. This explains the quartic growth rate of the group $H_{\mathbb Z}$. More generally, for any metric space the topological dimension is bounded above by the Hausdorff dimension, and so the finite-dimensionality of the limit space $Y$ follows from the polynomial growth of the group.