Revised
Interesting question.
Here's a thought:
You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms
$x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the
multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times
the order 2 group generated by $-1$.
If you take a submonoid that uses only
one prime, it is quasi-isometric to a quotient of the hyperbolic plane by an action of $\mathbb Z$, which is multiplication by $p$ in the upper half-space model. To see this, place a dots labeled by integer $n$ at position $(n*p^k, p^k)$ in the upper half plane, for every pair of integers $(n,k)$, and connect them by horizontal line segments and by vertical line segments whenever points are in vertical alignment. The quotient of upper half plane by the hyperbolic isometry $(x,y) \rightarrow (p*x, p*y)$ has a copy of the Cayley graph for this monoid. This is also quasi-isometric to the 1-point union of two copies of the hyperbolic plane, one for negative integers, one for positive integes. It's a fun exercise,
using say $p = 2$. Start from 0, and recursively build the graph by connecting $n$ to $n+1$ by one color arrow, and $n$ to $2*n$ by another color arrow. If you arrange positive integers in a spiral, you can make a neat drawing of this graph (or the corresponding graph for a different prime.) The negative integers look just the same, but with the successor arrow reversed.
If you use several primes, the picture gets more complicated. In any case, one can take rescaled limits of these graphs, based at sequence of points, and get asymptotic cones for the monoid. The graph is not homogeneous, so there is not just one limit.
Another point of view is to take limits of $\mathbb Z$ without rescaling, but
with a $k$-tuple of constants $(n_1, \dots , n_k)$. The set of possible identities among polynomials in $k$ variables is compact, so there is a compact space of limit rings for $\mathbb Z$ with $k$ constants. Perhaps this is begging the question: the identitites that define the limits correspond to diophantine equations that have infinitely many solutions.
Rescaling may eliminate some of this complexity.
A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to
$\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on some asymptotic cone for the affine monoid for $\mathbb Z$.
With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest.
Edit: I couldn't resist my predilection for generalizations: Using darij grinberg's simplification, the proof below shows:
Let $k$ be a field, $q \in GL_n(k)$ a matrix of finite exponent $m$ with char$(k) \nmid m$ and $M \subseteq k^n$. Futhermore, let $E$ be the eigenspace of $q$ corresponding to the eigenvalue $1$ and let $U \le k^n$ be the space spanned by the columns of $1-q$. Then the following is true for $A := 1+q+\dots + q^{m-1}$:
- $\lbrace x \in k^n \mid Ax \in M \rbrace = U + \frac{1}{m}(E \cap M)$
- $U$ and $(1/m)(E \cap M)$ intersect in $0$ iff $0 \in M$, otherwise the intersection is empty
- $A$ is diagonizable with diagonal $(m,...,m,0,...,0)$ where the number of m's equals $\dim E$
(Older formulation)
Let $E \le \mathbb{C}^n$ be the eigenspace of $1$ of the matrix $q$ and let $U \le \mathbb{C}^n$ be the space spanned by the columns of $1-q$.
Set $A := 1+q+\dots + q^{m-1}$ and $X:= \lbrace x \in \mathbb{C}^n \mid A\cdot x \in \mathbb{Z}^n \rbrace$ and $L := E \cap \mathbb{Z}^n$.
Then the following holds:
$X = U \oplus \frac{1}{m}L$.
Proof: Assume $\dim E = d$. Then $\dim U = \text{rank}(1-q) = n-d$.
Since each $x \in E$ satisfies $Ax = mx$, $E$ contains eigenvectors from $A$ of the eigenvalue $m$. From $A \cdot (1-q) = 0$ it follows that $U$ consists of eigenvectors of $A$ of the eigenvalue $0$. Hence $E \cap U = 0$ and for dimensional reasons
$$\mathbb{C}^n = U \oplus E.$$
Since $q$ has integral entries, it's possible to chosse a basis of $E$ in $\mathbb{Q}^n$ and by multiplying with a suitable integer it's also possible to choose a basis in $\mathbb{Z}^n$. Therefore $L = E \cap \mathbb{Z}^n$ is a lattice of rank $d$. Let $\lbrace e_1, \dots, e_d \rbrace$ be a basis of $L$. Let $x \in X$ and write
$$x = u + \sum_i \alpha_i e_i \text{ with } \alpha_i \in \mathbb{C}.$$
Then $Ax = \sum_i m\alpha_i e_i \in \mathbb{Z}^n$ and $q(Ax) = Ax$. It follows $Ax \in E \cap \mathbb{Z}^n = L = \oplus_i \mathbb{Z}e_i$ and therefore $m\alpha_i \in \mathbb{Z}$. This shows $X \subseteq U \oplus (1/m)L$. The converse inclusion is obvious. qed.
Edit: Also note that the image of $A$ is given by
$$ Y := \lbrace Ax \mid x \in X \rbrace = L.$$
Best Answer
Dima, I can not write a comment (yet) so I will start an answer to my own question.
You may assume that $\Gamma$ acts by isometries, so $A=\mathbb R^n/\Gamma$ is an Alexandrov space. For each maximal subgroup $F$ one can take its fixed point set $S_F$ in $\mathbb R^n$. $S_F$ is an (affine) subspace and image (say $E_F$) in $\mathbb R^n/\Gamma$ is a singular set (so called extremal subset of Alexandrov space). The maximality of $F$ implies that $E_F$ contains no proper extremal sets (a smaller subset is fixed by bigger group). (In fact $E_F$ is a flat manifold and its has a neighborhood which isometric for a product $E_F\times Cone$.)
So the question boils down to finding maximal number of such extremal sets in $A$. A particular case of such sets are isolated singular points. Perelman's theorem states that number of "one-point extremal sets" in an Alexandrov space with curvature $\ge 0$ is at most $2^n$. The proof repeats a proof of Erdős problem: if you have $m$ points in $\mathbb R^n$ such that all angles in all triangles $\le \pi/2$ then $m\le 2^n$. We take homothety with coefficient 1/2 for each point, then images of convex hull don't have common internal points (otherwise it would occur obtuse angle), then compairing volume of convex hull and its images gives estimate for number of points.