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Another possibility is to consider group elements rather than commutators. If you take the matrices $A=\mathrm{diag}(1,\xi,\ldots,\xi^{n-1})$ and $B=\begin{pmatrix}0&1&0&\cdots&0&0\\ 0&0&1&\cdots&0&0\\ 0&0&0&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&\cdots&0&1\\ 1&0&0&\cdots&0&0 \end{pmatrix}$ in $gl_n$ (here $\xi=\exp(2\pi i/n)$, then $A^n=B^n=I_n$, $BA=\xi AB$, and the elements $A^iB^j$ form a basis of $gl_n$ which has a nice multiplication table. Moreover, this way the algebra of matrices is $\mathbb{Z}_n\times\mathbb{Z}_n$-graded (put $\deg(A^iB^j)=(i,j)$, it works well). Considering various homomorphisms $\mathbb{Z}_n\times\mathbb{Z}_n\to\mathbb{Z}_n$, you will get all sorts of $\mathbb{Z}_n$-gradings.