After having identified $M_n(B(\mathcal{H}))$ with $B(\mathcal{H}^{(n)})$ one could take a unital $C^*$ Algebra $\mathcal{A}$ and identifiying $M_n(\mathcal{A}$) by representing it as
$$ \pi : M_n(\mathcal{A}) \rightarrow M_n(B(\mathcal{H})) $$
where $\pi$ is a $*-$ representation. Is it immediate then, to say that the image of $M_n(\mathcal{A})$ under $\pi$ is closed, so that $\pi(M_n(\mathcal{A}))$ is a $C^*$ Algebra?
EDIT:
I think I found out (a part) by myself:
If I consider two $C^*$ Algebras $\mathcal{A},\mathcal{B}$ and a $*-$ homomorphism between them $\pi$, then I know its image will be the same as
$$\tilde{\pi}: \mathcal{A}/ker(\pi) \rightarrow B $$
where $\mathcal{A}/ker(\pi)$ is a $C^*$ algebra (cause $ker(\pi)$ is a closed two-sided ideal) and $\tilde{\pi}$ is injective(hence isometric), so by homomorphism theorems $\mathcal{A}/ker(\pi) \cong Range(\pi)$. Thus $Range(\pi)$ is a $C^*$ Algebra (hence norm-closed).
Can I say the same thing for $M_n(\mathcal{A}$) and $M_n(\mathcal{B})$?
Without too much work you can show that $M_n(A)$ is closed. Namely, you can show that if $A$ is closed, then norm convergence in $M_n(A)$ is equivalent to entry-wise norm convergence in $A$. then, if $A$ is closed, so is $M_n(A)$.
Once you know that $M_n(A)$ is closed, your reasoning applies.