The answer is no, even for $n=2$.
One thng to note is that over the decades, different groups of authors have used different terminology. The property you call Dedekind-finite has been called directly finite, or von Neumann finite. Then given a unital ring $R$, people have asked when $M_n(R)$ is DF for all $n\geq 1$, and this property has been called weakly finite (I believe this terminology was introduced by P. M. Cohn) or fully von Neumann finite, or (in the context of ${\rm C}^*$-algebras) stably finite. I mention all this in case it helps to look things up in the literature.
It turns out that one can find a unital ring $R$ which has no non-trivial zero-divisors (so in particular no non-trivial idempotents, which in turn implies $R$ is Dedekind finite) yet has the property that $M_2(R)$ is not Dedekind finite. See
J. C. Shepherdson, Inverses and zero divisors in matrix rings.
Proc. London Math. Soc. (3) 1, (1951). 71–85. MR0041831
which was pointed out to me by Mr A. Fisher.
I can't resist pointing out another example. Much later, it was shown that there is a unital, finite ${\rm C}^*$-algebra $A$ (which can be realised as an extension of $C({\mathbb T}^3)$ by the compacts) such that $M_k(A)$ is not DF for some $k\geq 2$; however, at a quick glance I don't see how to obtain an actual value of $k$ from the proof. The paper is:
N. P. Clarke, A finite but not stably finite ${\rm C}^*$-algebra.
Proc. Amer. Math. Soc. 96 (1986), 85–88 | MR0813816