On $M_n(\mathbb{C})$, if we take the operator norm by acting on $\mathbb{C}^n$ where $||(z_1,\ldots,z_n)||=\max_{1\leq i\leq n}||z_i||$, then for $[a_{ij}]\in M_n(\mathbb{C})$ we have $||[a_{ij}]||=\max_i \sum_j ||a_{ij}||$.
For a general Banach algebra $A$, if we take the operator norm by acting on $(A^+)^n$ where $A^+$ is the unitization of $A$ and $||(b_1,\ldots,b_n)||=\max_i ||b_i||$ for $(b_1,\ldots,b_n)\in (A^+)^n$, do we still have $||[a_{ij}]||=\max_i \sum_j ||a_{ij}||$ for $[a_{ij}]\in M_n(A)$? The inequality $\leq$ seems fine but I don't see equality.
(I'm taking $(A^+)^n$ here but is there a difference if I take $A^n$?)
Let $\lambda_a \colon b \mapsto a\cdot b$. Then it's easy to see that the operator norm of the matrix is at most $$\max_i \sum_j \lVert \lambda_{a_{ij}}\rVert_{\operatorname{op}}.$$ For a general Banach algebra, it is possible that $\lVert \lambda_a \rVert_{\operatorname{op}} < \lVert a\rVert$ for some $a$.
Still, even if we take the operatornorm of the left multiplication (a no-op if the two norms agree), for $n > 1$ the inequality is in general not an equality. Consider $A = C(K)$ for a nice compact space $K$. Then $A$ is even a unital $C^\ast$-algebra. But if the $a_{ij}$ have disjoint support, then
$$\max \Biggl\{ \Biggl\lVert \sum_{j = 1}^n a_{ij}\cdot b_j\Biggr\rVert : \lVert b_j\rVert \leqslant 1 \text{ for } 1 \leqslant j \leqslant n\Biggr\} = \max_j \lVert a_{ij}\rVert.$$