Is $m=n+1$ the largest $m$ such that $S_m$ has a faithful action on $\mathbb{Z}^n$? 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.)
 A: Here's a reasonably elementary argument that shows that $m \le 2n+3$. By Bertrand's postulate there's a prime $p$ between $n+2$ and $2n+3$. Then $S_p$ contains an element of order $p$. characteristic polynomial of an integer matrix of order $p$ is necessarily divisible by the cyclotomic polynomial $\Phi_p(x) = x^{p-1} + ... + 1$, which is irreducible, so if such a matrix is $n \times n$ then we conclude that $p-1 \le n$, which contradicts $p \ge n+2$. Hence $S_p$ can't embed into $\text{GL}_n(\mathbb{Z})$. 
(Asymptotically we actually expect that there is a prime between $n$ and $n + O(\sqrt{n})$ so $m$ gets reasonably close to $n$ for large $n$.) 
Here's a silly argument that shows that $m \le n+6$ for $n \ge 2$. Conditional on Goldbach's conjecture (never thought I'd see myself write that!), one of $\{ n+3, n+4, n+5, n+6 \}$ is a sum of two distinct primes $p$ and $q$. Write $m = p+q$. Thus $S_m$ contains the product of a cycle of order $p$ and a cycle of order $q$. The characteristic polynomial of an integer matrix with this property is divisible by the product $\Phi_p(x) \Phi_q(x)$, so if such a matrix is $n \times n$ then we conclude that $p + q - 2 \le n$, which contradicts $p + q \ge n+3$. Hence $S_m$ can't embed into $\text{GL}_n(\mathbb{Z})$.
According to Wikipedia, the smallest irreducible representations of $S_m$ of dimension greater than $1$ have dimension at least $m-1$ for $m \ge 7$, so your conjecture is correct for all $n \ge 5$. In fact this result shows that $S_{n+2}$ cannot embed into $\text{GL}_n(\mathbb{C})$, let alone $\text{GL}_n(\mathbb{Z})$. 
