Yes, for odd $n$, the ring $(\mathbb Z/n)[i,j,k]$ is isomorphic to the ring of two-by-two matrices over $\mathbb Z/n$.
To explain this, it is better to write $\mathbb Z/n$ and $\mathbb Z/p$ rather than $\mathbb Z_n$ and $\mathbb Z_p$, because, in fact, the decisive statement concerns $p$-adic integers $\mathbb Z_p$, and we should reserve the notation for that.
What is true is that, using the notation for $p$-adic integers, for odd prime $p$, $\mathbb Z_p[i,j,k]$ is the full matrix ring over the $p$-adic integers $\mathbb Z_p$. Granting this for a moment, $(\mathbb Z/p^\ell)[i,j,k]$ is the image of $\mathbb Z_p[i,j,k]$ by mapping $p^\ell \mathbb Z_p$ to $0$, since $\mathbb Z_p/p^\ell \mathbb Z_p\approx \mathbb Z/p^\ell \mathbb Z$.
By Sun-Ze's theorem, the general $\mathbb Z/n$ is the sum of the $\mathbb Z/p^\ell$ where $p^\ell$ are the prime powers dividing $n$. Thus, for odd $n$, $(\mathbb Z/n)[i,j,k]$ is the sum of the corresponding rings for the $p^\ell$.
To prove that $\mathbb Z_p[i,j,k]$ is the full matrix ring, start with the point that $\mathbb Q_p[i,j,k]$ is the full matrix ring, for odd $p$. There are various proofs of this... For example, the surjectivity of norms on finite fields, together with Hensel's Lemma, proves that the quadratic form $a^2+b^2+c^2+d^2=\hbox{norm}(a+bi+cj+dk)$ has a non-trivial zero, so $\mathbb Q_p[i,j,k]$ is not a division ring.
The additional ingredient is that the ring of "Hurwitz integers" consisting of $\mathbb Z[i,j,k]$ with $(1+i+j+k)/2$ adjoined is Euclidean, in the appropriate sense for a non-commutative ring. "Locally" at odd primes $p$, the $1/2$ in the definition of the Hurwitz integers is a unit, so, locally, at odd primes, the $p$-adic version of the Hurwitz integers is the naive notion of local quaternion integers, $\mathbb Z_p[i,j,k]$. The rest of the discussion is just clean-up.
