Is $R^n$ projective as a $M_n(R)$-module?

Whenever $e$ is an idempotent element in a ring $S$, the left ideal $Se$ of $S$ is, when viewed as a module, projective. Indeed, one can easily check that $f=1-e$ is also idempotent and that $S=Se\oplus Sf$.

Now if $S=M_n(R)$ is a matrix ring over some other ring $R$, the elementary matrix $e=E_{1,1}$ (the one whose components are all zero except for the one in the position $(1,1)$, which is equal to $1$) is idempotent. You should check that the left ideal $Se$ is isomorphic to $R^n$ as an $S$-module.

Note that $M_n(R) \cong (R^{n})^{\oplus n}$ by considering a matrix's columns as tuples in $R^n$. (This is an isomorphism of $M_n(R)$-modules by the definition of matrix multiplication: to multiply two matrices $A$ and $B$, we apply $A$ to each of the columns of $B$.) Then $R^n \oplus (R^n)^{\oplus n-1} \cong M_n(R)$, so $R^n$ is a direct summand of a free module, hence is projective.