Given a non-unital $C^*$-algebra $B$, it is proved in this post that $\mathcal{M}(B)$, the multiplier algebra of $B$ coincides with the closure of the strict topology of $B$ where the strict topology is the locally convex topology generated by the following family of semi-norms: for each $b\in B$:
$$ \rho_b: B\rightarrow [0, \infty), \hspace{0.3cm} x\mapsto \|\,xb\,\| $$ Using the definition of the strict topology, it can be shown that, for each $n\in\mathbb{N}$, $M_n\big[\mathcal{M}(B)\big] = \mathcal{M}\big[ M_n(B) \big]$ (after we embed $B$ into some $B(H)$) and it is proved in Proposition 2.2.11, K-Theory and $C^{\ast}$-Algebras written by N.E. Wegge-Olsen that the multiplier algebra of a $C^*$-algebra is independent to the choice of $B(H)$. In the same proposition, it is also shown that $\mathcal{M}(B)$ is isometric isomorphic to the space of double centralizers of $B$ and hence $\mathcal{M}(A\bigoplus B)\cong \mathcal{M}(A)\bigoplus\mathcal{M}(B)$ for an arbitrary $C^*$-algebra $A$.
If statements above are correct, then I can conclude that, given a fintie-dimensional $C^*$-algebra $A$, I will have $\mathcal{M}(A\bigotimes_{\min}B)\cong A\bigotimes_{\min}\mathcal{M}(B)$. Could anyone provide another example of unital $C^*$-algebra $A$ that satisfies this isomorphism, or prove that, to have this isomorphism $A$ necessarily needs to be finite-dimensional? What mainly blocks me is that I do not know in general when the minimal tensor norm coincides with the strict topology.
Posting this as a community wiki answer so that the question is filed under answered:
The following is a theorem of Akemann, Pedersen and Tomiyama:
The theorem addresses more general $C^*$-norms on the tensor products and can be found in the paper "Multipliers of $C^*$-algebras" (here).
See also this related post on Math Overflow.