Is $PSL(n, \mathbb Z)$ isomorphic to a subgroup of $GL(n,\mathbb C)$ or even $GL(n+1,\mathbb C)$?

You can look for a finite subgroup of $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for even $n\ge 6$. In $GL(n,\mathbb Z)$ take the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$. $G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n2}$. In any faithful action of $G$ on $\mathbb C^d$ there is a onedimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$dimensional characters of $D$ every orbit has at least $n(n1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n1)/2>n$. One step breaks down when $n=4$, so you'd need a different subgroup in that case. 


If $n \ge 3$, perhaps one can use Margulis superrigidity to deduce that any homomorphism from $\mathrm{SL}(n,\mathbb{Z})$ to $\mathrm{GL}(m,\mathbb{C})$ extends to a representation of $\mathrm{SL}(n,\mathbb{C})$. The standard representation of $\mathrm{SL}(n,\mathbb{Z})$ factors through $\mathrm{PSL}(n,\mathbb{Z})$ if and only if $n$ is odd. The next smallest representation of $\mathrm{SL}(n,\mathbb{C})$ has dimension $> n+1$, (assuming that $n > 2$). If $n = 2$, then $\mathrm{PSL}(2,\mathbb{Z})$ is certainly a subgroup of $\mathrm{GL}(3,\mathbb{Z})$. So the answer seems to be: if and only if $n = 2$ or $n$ is odd. 

