Question: Let $A$ and $B$ be $C^\*$-algebras. If $A$ is finite-dimensional, then I want to show that the algebraic tensor product $A \odot B$ is complete for the minimal $C^\*$-norm on it (in fact, the choice of norm does not matter since $A$ is nuclear).

**Attempt:**

I know that $$A \cong M\_{n\_1}(\mathbb{C}) \oplus \dots \oplus M\_{n\_k}(\mathbb{C})$$

And thus
$$A \odot B \cong (A \odot M\_{n\_1}(\mathbb{C}))\oplus \dots \oplus (A\odot M\_{n\_k}(\mathbb{C})) $$

so I believe it suffices to show that $A \odot M\_{n\_1}(\mathbb{C})\cong A \odot \mathbb{C}^{n^ 2}$ is complete.

Then I'm stuck.

Maybe I'm overcomplicating this though....

Answer: 
You wrote as if $B$ is the finite-dimensional one, so I'll stick with that.

Let $\gamma$ be a C$^\*$-norm on $A\otimes M\_n(\mathbb C)$ (one certainly exists, because we can represent $A\subset B(H)$ and then $A\otimes M\_n(\mathbb C)$ can be represented in $B(H\otimes\mathbb C^n)$).

For any $k$, the map $a\longmapsto \gamma(a\otimes E\_{kk})$ defines a C$^\*$-norm on $A$. As a C$^\*$-algebra admits a unique C$^\*$-norm, we get that $\gamma(a\otimes E\_{kk})=\|a\|$. Even if $k\ne j$,
$$
\gamma(a\otimes E\_{kj})^2=\gamma(a^\*a\otimes E\_{jj})=\|a^\*a\|=\|a\|^2.
$$
Now suppose that we have a Cauchy sequence $\{x\_n\}\subset A\otimes M\_n(\mathbb C)$. We can write $$x\_n=\sum\_{k,j} a\_n(k,j)\otimes E\_{k,j}.$$
For any $k,j$,
\begin{align}
\|a\_n(k,j)-a\_m(k,j)\|^2&=\gamma((a\_n(k,j)-a\_m(k,j))\otimes E\_{11})\\[0.3cm]
&=
\gamma((1\otimes E\_{11})(x\_n-x\_m)(1\otimes E\_{11}))\\[0.3cm]
&\leq\gamma(x\_n-x\_m).
\end{align}
So each sequence $\{a\_n(k,j)\}\_n$ is Cauchy on $A$. As $A$ is a C$^\*$-algebra, there exist $a(k,j)\in A$ with $\lim\_na\_n(k,j)=a(k,j)$. We want to show that $x\_n\to\sum\_{k,j} a(k,j)\otimes E\_{k,j}$. For this
\begin{align}
\gamma(x\_n-\sum\_{k,j} a(k,j)\otimes E\_{k,j})
&=\gamma(\sum\_{k,j} (a\_n(k,j)-a(k,j))\otimes E\_{k,j})\\[0.3cm]
&\leq \sum\_{k,j} \gamma((a\_n(k,j)-a(k,j))\otimes E\_{k,j})\\[0.3cm]
&=\sum\_{k,j} \|a\_n(k,j)-a(k,j)\|\\[0.3cm]
&\to0,
\end{align}
where it is essential in this last step that the sum is finite. So $A\otimes M\_n(\mathbb C)$ is complete with respect to $\gamma$, and so it is a C$^\*$-algebra. This in particular shows that $\gamma$ is necessarily the norm described in the second paragraph.
The point is that $A \odot B$ is closed in $A \otimes B$. To see this, choose a basis $\{e\_i\}\_{i=1}^n$ for $A$ and suppose that $\{x\_k\}\_{k=1}^\infty$ is a sequence in $A \odot B$ that converges to $x \in A \otimes B$. Then we can write
$$x\_k = \sum\_{i=1}^n e\_i \otimes b\_i^k$$
and since $x\_k \to x$ in $A \otimes B$, applying the slice map $\tau\_j \otimes \iota: A \otimes B \to B$ where $\tau \in A^\*$ satisfies $\tau(e\_i) = \delta\_{i,j}$, we see that
$$b\_j^k = (\tau\_j \otimes \iota)(x\_k) \to (\tau\_j \otimes \iota)(x).$$
Hence, $$x=\sum\_{i=1}^n e\_i \otimes (\tau\_i \otimes \iota)(x) \in A \odot B.$$