Suppose $\mathcal{H}$ is a complex Hilbert space and $\mathfrak{A}$ is a von Neumann algebra of operators on $\mathcal{H}$. Then $M_n(\mathfrak{A})$ can be thought of an algebra (multiplication is matrix multiplication) of operators on $\mathcal{H}^{(n)}:=\mathcal{H}\oplus\dots\oplus\mathcal{H}$ ($n$ times). I have managed to prove that (w.r.to operator norm of $\mathcal{B}(\mathcal{H}^{(n)})$), $M_n(\mathfrak{A})$ forms an von Neumann algebra.
What is the commutant of $M_n(\mathfrak{A})$ in $\mathcal{B}(\mathcal{H}^{(n)})$?
It is easy to see that the diagonal matrices with entries from $\mathfrak{A}'$ lie in $M_n(\mathfrak{A})'$: For if $T\in \mathfrak{A}'$ and $T^{(n)}:=T\oplus\dots\oplus T$ then for any $A=[A_{ij}], A_{ij}\in \mathfrak{A}$ we have $\forall i, j, A_{ij}T=TA_{ij}\implies AT^{(n)}=T^{(n)}A$.
Therefore my claim is:
$M_n(\mathfrak{A})'=\{T^{(n)}:T\in \mathfrak{A}'\}$; where $T^{(n)}:=T\oplus\dots\oplus T$ ($n$ times)
For this take $T\in M_n(\mathfrak{A})'$. Write $T=[T_{ij}]$ where $T_{ij}\in \mathcal{B}(\mathcal{H})$ (This can be done always since $T\in \mathcal{B}(\mathcal{H}^{(n)})$)
Consider the matrices $E_{ij}\in M_n(\mathfrak{A})$, whose $(i,j)$th entry is $I\in \mathfrak{A}$ and rest is $0$. Then since $T$ commutes with all of these matrices then simple matrix multiplication will show that $T$ is diagonal with equal diagonal entries, i.e., $T=A\oplus\dots\oplus A$ ($n$ times) with $A \in \mathfrak{A}$.
Also if $B\in \mathfrak{A}$ and $S:=B\oplus\dots\oplus B$ ($n$ times), then $ST=TS$ which implies $AB=BA$, i.e., $A\in \mathfrak{A}'$.
Is the proof alright?
Edit. From the above description of $M_n(\mathfrak{A})'$ now it is easy to calculate the double commutant $M_n(\mathfrak{A})''$. Indeed $$ M_n(\mathfrak{A})''=M_n(\mathfrak{A}'') $$ For if $B=[B_{ij}]\in M_n(\mathfrak{A})''$ then $BA^{(n)}=A^{(n)}B$ for all $A\in \mathfrak{A}'$. But this implies $\forall i, j, ~ B_{ij}A=AB_{ij}$ for all $A\in \mathfrak{A}'$, i.e., $\forall i, j~B_{ij}\in \mathfrak{A}''$.
This last result indeed proves (using von Neumann Double Commutant theorem): $M_n(\mathfrak{A})$ is a von Neumann algebra if $\mathfrak{A}$ is so!
Yes, this proof is alright. More generally, the commutant of the tensor product $\mathfrak M\overline{\otimes} \mathfrak N$ is $\mathfrak M'\overline{\otimes}\mathfrak N'$ (Theorem IV.5.9 in Takesaki), which yields $$ M_n(\mathfrak M)'=(M_n(\mathbb C)\overline{\otimes}\mathfrak M)'=\mathbb CI_n\overline{\otimes}\mathfrak M'. $$ But in this specific case your proof is simpler (and one still would have to prove $M_n(\mathbb C)'=\mathbb C I_n$, which is almost as hard as the original statement).