Let $D$ be a skew field, and consider an arbitrary noncommutative unital ring $R$ extending the matrix ring $D^{n\times n}$. Must $R$ be free as a $D^{n\times n}$-module?
If $R$ is finite-dimensional, then $R$ is indeed free, as I pointed out in the comments to this question (which inspired this one). Likewise if $n=1$, by undergraduate linear algebra. But is the same true for infinite-dimensional rings when $n\geq2$?
Now, $\mathbb{Z}^{\mathbb{Z}}$ is not free over $\mathbb{Z}$. So the simplest example I can think of is the infinite direct product $(\mathbb{C}^{2\times 2})^{\mathbb{Z}}$ over $\mathbb{C}^{2\times 2}$. But I can't tell if even such a simple example is free.
Surprisingly, the answer is "yes, all such rings are free." (Your proposed counterexample is not one.)
To see this, first note that $D$ and $D^{n\times n}$ are Morita-equivalent (Corollaries 4.5 and 5.5 of these notes). Since $D$-vector spaces are projective (they're free!), every $D^{n\times n}$-module must be projective too. But then Corollary 3.2 of Hyman Bass, "Big Projective Modules are Free" proves that any infinitely-generated projective module over a simple Noetherian ring (such as $D^{n\times n}$) is free.
I do not know any elementary argument to prove this result.