First, it is easy to see that $\mathrm{Comm}(\mathbb{Z}^n)$ must be isomorphic to a subgroup of $\mathrm{GL}(n,\mathbb{Q})$. In particular, every finite-index subgroup of $\mathbb{Z}^n$ is an $n$-dimensional lattice, so any isomorphism between two such subgroups extends uniquely to an isomorphism $\mathbb{Q}^n\to\mathbb{Q}^n$. Note that two isomorphisms of $\mathbb{Q}^n$ that agree on an $n$-dimensional lattice must be equal, so it really does work to think of elements of $\mathrm{Comm}(\mathbb{Z}^n)$ as matrices.
All that remains is to show that every element of $\mathrm{GL}(n,\mathbb{Q})$ corresponds to some element of $\mathrm{Comm}(\mathbb{Z}^n)$. This is fairly easy: if $A\in\mathrm{GL}(n,\mathbb{Q})$, then there exists a positive integer $k$ so that the $n\times n$ matrix for $kA$ has integer entries. In this case, $A$ maps the finite-index subgroup $k\\,\mathbb{Z}^n$ (of index $k^n$) isomorphically to another finite-index subgroup of $\mathbb{Z}^n$. The image subgroup has finite index since it spans $\mathbb{Q}^n$, and is therefore an $n$-dimensional lattice.
Edit: As Mark points out, this is essentially the same answer given in the comments above.