It's possible for any $n$ divisible by 4, other than $n=8$.
First, find a magic square of size $n/4$. (This is impossible when $n=8$.) Call it $\mathcal{M}$. Now consider the following 4 by 4 magic square $\mathcal{S}$, using the numbers 0 through 15:
$\left( \begin{array}{ccc}
8 & 5 & 2 & 15 \\
3 & 14 & 9 & 4 \\
13 & 0 & 7 & 10 \\
6 & 11 & 12 & 1 \end{array} \right)$
For each number $k$ in this square, replace it by a copy of $\mathcal{M}$ with every number increased by $kn^2/16$. One can easily check that the resulting square must be magic, because both of the starting squares are.
If you've heard of a tensor product or Kronecker product, then this should feel like the product of two magic squares.