Let me elaborate on Johannes Ebert's comment.
Since you are dealing with commuting operators, you can find a basis for $V$ so that the matrices for $A_1,\ldots,A_n$ are in Jordan normal form. So, $V$ decomposes as $V=V_1\oplus\cdots\oplus V_r$, where for each $i=1,\ldots,r$, there exists scalars $a_{ij}$, $j=1,\ldots,n$, such that $(x_j-a_{ij})^Nv=0$ for all $v\in V_i$ and $N\gg0$. Moreover, we may assume that each $V_i$ is indecomposable.
Now, we may assume $r=1$, and let $a_j$ be the generalized eigenvalues for the action of $x_j$ on $V$. By replacing $x_j$ by $x_j-a_j$, we may assume all $a_j$ are 0, and the $x_j$ are nilpotent.
Now, the problem is to identify commuting families of $n$ nilpotent operators acting on a finite dimensional vector space, say of dimension $N$. To do this, for each subset $S\subset\{ 1,\ldots,N-1\}$, consider the matrix $e_S=\sum_{i\in S}e_{i,i+1}$. Then, we are looking for a collection of subsets $S_1,\ldots,S_n$ such that $e_{S_i}e_{S_j}=e_{S_j}e_{S_i}$. If we consider the special case where $S_i$ and $S_j$ are intervals, we see that we must have either $S_i=S_j$ or $S_i\cup S_j$ consists of two disjoint intervals. To generalize, each $S_i=\bigcup S_{ik}$, where each $S_{ik}$ is an interval, but no $S_{ik}\cup S_{il}$ is an interval (or $S_i=\emptyset$).