Nuclearity of finite dimensional $C^*$-algebras

I'll do $M_n(\mathbb C)$ for simplicity, as any other finite-dimensional C$^*$-algebra is just a direct sum of those.

If one defines nuclearity in the sense that the identity is a nuclear map, the statement is completely trivial. So let us focus on the case where we define nuclearity as "unique norm on any tensor product".

The key here is to notice that $M_n(\mathbb C)\odot A\simeq M_n(A)$ is a C$^*$-algebra; so there is no completion to be taken. The isomorphism is trivial, as one maps $\sum_{kj}E_{kj}\otimes a_{kj}\longmapsto \big[a_{kj}\big]$. We can represent $M_n(A)\subset B(H^n)$, for a representation $A\subset B(H)$. So all one needs is to show that $M_n(A)$ is complete in $B(H^n)$.

Let $\{R_s \}_s \subset M_n(A)$ be a Cauchy sequence. Write $R_s =[a_{kj,s }]$. Given any $\xi,\eta\in H$, $$ |\langle a _{kj,s }\xi,\eta\rangle|=\Bigg|\Bigg\langle R_s \begin{bmatrix} \xi\\ 0\\ \vdots\\ 0\end{bmatrix},\begin{bmatrix} \eta\\ 0\\ \vdots\\ 0\end{bmatrix}\Bigg\rangle\Bigg|\leq\|R_s \|\,\|\xi\|\,\|\eta\|. $$ It follows that $\|a_{kj,s }\|\leq\|A_s \|$. As the sequence $\{A_s\}$ is Cauchy, we deduce that each entry $\{a_{kj,s}\}$ forms a Cauchy sequence. By the completeness of $A$, there exists $a_{kj}\in A$ with $a_{kj}=\lim_sa_{kj,s}$. Let $R=[a_{kj}]\in M_n(A)$. Fix $\varepsilon>0$; there exists $s_0$ such that for all $s\geq s_0$, $\|a_{kj,s}-a_{kj}\|<\varepsilon$ for all $k,j$. Then, for $\tilde\xi\in H^n$, with $\tilde\xi=(\xi_1,\ldots,\xi_m)^T$, \begin{align} \|(R_s-R)\xi\|^2&=\sum_{k,j,\ell}\langle (a_{\ell k,s}^*-a_{\ell k}^*)(a_{\ell j,s}-a_{\ell j})\xi_j,\xi_k\rangle \leq \varepsilon^2\,n\,\sum_{k,j}\|\xi_j\|\,\|\xi_k\|\\[0.3cm] &=\varepsilon^2\,n\,\Big(\sum_{k}\|\xi_j\|\Big)^2 \leq \varepsilon^2\,n^2\,\sum_{k}\|\xi_j\|^2=\varepsilon^2\,n^2\,\|\xi\|^2. \end{align} Thus $R_s\to R$ and $M_n(A)$ is complete.