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This answer arose by a discussion with @JamieGabe in the comments.

One can prove that the map $$\Phi: M_n(A) \to M_n(A): A \mapsto n \operatorname{Diag}(A)-A$$ is completely positive [Paulsen, "Completely bounded maps and operator spaces", exercise 3.6].

In particular, it is positive. Hence, writing $A= (a_{i,j})$ as in the OP, we obtain $$A \le n \operatorname{Diag}(A)$$ and taking norms leads to $$\|A\| \le n \|\operatorname{Diag}(A)\|=n \max_{i=1}^n \|a_{i,i}\| \le n \left\|\sum_{i=1}^n a_{i,i}\right\|.$$

This answer arose by a discussion with @JamieGabe in the comments.

One can prove that the map $$\Phi: M_n(A) \to M_n(A): A \mapsto n \operatorname{Diag}(A)-A$$ is completely positive [Paulsen, "Completely bounded maps and operator spaces", exercise 3.6].

In particular, it is positive. Hence, writing $A= (a_{i,j})$ as in the OP, we obtain $$A \le n \operatorname{Diag}(A)$$ and taking norms leads to $$\|A\| \le n \lVert\operatorname{Diag}(A)\rVert=n \max_{i=1}^n \|a_{i,i}\| \le n \left\|\sum_{i=1}^n a_{i,i}\right\|.$$

Following @JamieGabe's hint, we claim that

$$\lVert(a_{i,j})\Vert \le n \Bigl\lVert\sum_i a_{i,i}\Bigr\rVert.$$

Indeed, the map

$$\operatorname{Tr}: M_n(A) \to A: (a_{i,j})\mapsto \sum_i a_{i,i}$$ is easily checked to be completely positive.

Let $\{u_i\}$ be an approximate unit for $A$. Then the matrices $$U_i:= \begin{pmatrix}u_i & 0 & \cdots & 0\\ 0 & u_i & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\cdots & u_i\end{pmatrix} \in M_n(A)$$ determine an approximate unit for $M_n(A)$. By general properties of completely positive maps, $$\lVert\operatorname{Tr}\rVert = \lim_i \lVert\operatorname{Tr}(U_i)\rVert = \lim_i \lVert\nu_i\rVert = n$$ and the desired inequality follows.

This answer arose by a discussion with @JamieGabe in the comments.

One can prove that the map $$\Phi: M_n(A) \to M_n(A): A \mapsto n \operatorname{Diag}(A)-A$$ is completely positive [Paulsen, "Completely bounded maps and operator spaces", exercise 3.6].

In particular, it is positive. Hence, writing $A= (a_{i,j})$ as in the OP, we obtain $$A \le n \operatorname{Diag}(A)$$ and taking norms leads to $$\|A\| \le n \|\operatorname{Diag}(A)\|=n \max_{i=1}^n \|a_{i,i}\| \le n \left\|\sum_{i=1}^n a_{i,i}\right\|.$$

Following @JamieGabe's hint, we claim that

$$\lVert(a_{i,j})\Vert \le n \Bigl\lVert\sum_i a_{i,i}\Bigr\rVert.$$

Indeed, the map

$$\operatorname{Tr}: M_n(A) \to A: (a_{i,j})\mapsto \sum_i a_{i,i}$$ is easily checked to be completely positive.

Let $\{u_i\}$ be an approximate unit for $A$. Then the matrices $$U_i:= \begin{pmatrix}u_i & 0 & \cdots & 0\\ 0 & u_i & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\cdots & u_i\end{pmatrix} \in M_n(A)$$ determine an approximate unit for $M_n(A)$. By general properties of completely positive maps, $$\lVert\operatorname{Tr}\rVert = \lim_i \lVert\operatorname{Tr}(U_i)\rVert = \lim_i \lVert\nu_i\rVert = n$$ and the desired inequality follows.

Following @JamieGabe's hint, we claim that

$$\lVert(a_{i,j})\Vert \le n \Bigl\lVert\sum_i a_{i,i}\Bigr\rVert.$$

Indeed, the map

$$\operatorname{Tr}: M_n(A) \to A: (a_{i,j})\mapsto \sum_i a_{i,i}$$ is easily checked to be completely positive.

Let $\{u_i\}$ be an approximate unit for $A$. Then the matrices $$U_i:= \begin{pmatrix}u_i & 0 & \cdots & 0\\ 0 & u_i & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\cdots & u_i\end{pmatrix} \in M_n(A)$$ determine an approximate unit for $M_n(A)$. By general properties of completely positive maps, $$\lVert\operatorname{Tr}\rVert = \lim_i \lVert\operatorname{Tr}(U_i)\rVert = \lim_i \lVert\nu_i\rVert = n$$ and the desired inequality follows.