Is $R^n$ projective as a $M_n(R)$-module? It is clear that $R^n$ is not free over $M_n(R)$. But is it projective?
I suspect that it should be projective because we can probably come up with a projective basis, but I'm not sure how to find the basis. 
Moreover, ideally if $R^n$ were projective over $M_n(R)$, we would get a big class of projective modules that are not free which would be interesting, I guess.

Yes: $R^n$ is a finitely generated projective generator of $\textrm{Mod-}R$, so it is a finitely generated projective generator as a module over its endomorphism ring, which is the ring of matrices $M_n(R)$.
This is quite easy in general. Let $P_R$ be a finitely generated projective generator of $\mathrm{Mod\text{-}}R$ and let $S=\operatorname{End}(P_R)$. Then $P$ is a left $S$-module. Let's prove it is a finitely generated projective generator.
Consider a (split) epimorphism $R^n\to P$. By applying $\operatorname{Hom}_R(-,P)$, we get the split monomorphism
$$
\operatorname{Hom}_R(P,P)\to\operatorname{Hom}_R(R^n,P)
$$
The domain is isomorphic to $S$ as a left module, the codomain is isomorphic to $P^n$ as $S$-modules. Thus $S$ is a direct summand of ${}_SP^n$ and so ${}_SP^n$ is a generator of $S\textrm{-Mod}$, which implies ${}_SP$ is a generator as well.
Since $P_R$ is a generator, there is a split epimorphism $P^n\to R$. Then, applying $\operatorname{Hom}_R(-,P)$, we get a split monomorphism $\operatorname{Hom}_R(R,P)\to\operatorname{Hom}_R(P^n,P)$. The domain is isomorphic to ${}_SP$ and the codomain is isomorphic to ${}_SS^n$. Thus ${}_SP$ is a direct summand of ${}_SS^n$ and therefore it is finitely generated projective.
