I can write down a faithful action of $S_{n+1}$ on $\mathbb{Z}^n$. That is, I know of a way to explicitly give a homomorphism from $S_{n+1}$ to $GL(n,\mathbb{Z})$ that has a trivial kernel. An example of this is below, although I'm sure there are many others.
Can this ever be done with $S_{n+2}$, maybe for the right value of $n$? Are there larger symmetric groups than $S_{n+1}$ that can act faithfully on $\mathbb{Z}^n$ if $n$ has the right value? Or will $S_{n+2}$ never act faithfully on $\mathbb{Z}^n$?
The orders of maximal finite subgroups of $GL(n,\mathbb{Z})$ for $n=2,3,4,5$ that are tabulated here imply that for these $n$ values at least, the answer is negative, since $(n+2)!$ never divides the order of subgroups.
(One example of an $S_{n+1}$-action is to identify $\mathbb{Z}^n$ with polynomials of degree $n-1$ or less, permute the $n+1$ coefficients of $p(x)\cdot(x-1)$, and then divide by $(x-1)$ which will still be a factor after the permuting.)