Definition: Norm $p$ on $\mathbb C^{n\times n}$ is consistent if $p(AB)\leq p(A)p(B)$ for all $A,B\in \mathbb C^{n\times n}$.
I want to show
$p(C)=\max_i \sum_j |c_{ij}|$ is consistent, where $C\in \mathbb C^{n\times n}$.
By verifying the definition directly, I got a page of messy expressions, which did not lead me to anywhere. Also, Cauchy-Schwarz does not work well here. Any hint will be appreciated.
Let's write $q(C)=\max\{\|Cx\|_\infty:x\in\mathbb C^n,\|x\|_\infty=1\}$. I will show that $q(C)=p(C)$.
If $x\in\mathbb C^n$ and $\|x\|_\infty=1$, we have $$\|Cx\|_\infty=\max_i\left|\sum_jc_{ij}x_j\right|\leq\max_i\sum_j|c_{ij}||x_j|\leq\max_i\sum_j|c_{ij}|\cdot\|x\|_\infty=p(C).$$ Taking the supremum over such $x$ yields $q(C)\leq p(C)$.
Now choose $i$ so that $p(C)=\sum_j|c_{ij}|$, and consider the vector $x=(e^{-i\arg(c_{i1})},\ldots,e^{-i\arg(c_{in})})\in\mathbb C^n$. Then $\|x\|_\infty=1$, and
$$p(C)=\sum_j|c_{ij}|=\sum_jc_{ij}e^{-i\arg(c_{ij})}=\left|\sum_jc_{ij}x_j\right|\leq\|Cx\|_\infty\leq q(C).$$ Therefore $p(C)=q(C)$.
Now it's easy to see the norm $q$ is, as you say, consistent. For if $x\in\mathbb C^n$ and $\|x\|_\infty=1$, then $$\|ABx\|_\infty\leq q(A)\|Bx\|_\infty\leq q(A)q(B)\|x\|_\infty=q(A)q(B),$$ and taking the supremum over $x$ yields $q(AB)\leq q(A)q(B)$.