The matrix L_p norm means $\max\limits_{\|x\|_p = 1} \|Ax\|_p$, here $x$ is an $n \times 1$ vector, and the $Ax$ is an $n \times 1$ vector too. so when $p=2$, it is the well known operator norm.
Thanks. I try to answer the question in a particular sense. It can be seen that even for the operator norm, that is $p=2$, when n is really large, a entry wise random $+1,-1$ Bernoulli matrix have the largest singular value similar to $\sqrt{n}$, but the absolute matrix have the largest singular value $n$, so asymptotically, when $n$ is really large, we cannot have a constant $C>0$ such that $C \cdot \|A \|_2 \geq \|A' \|_2$.
As @Pietro Majer has said, I think it is easy to get $\|A \|_p \leq \|A' \|_p$. but the other way around for a fixed n is still hard to me.
The matrix L_p norm means $\max\limits_{\|x\|_p = 1} \|Ax\|_p$, here $x$ is an $n \times 1$ vector, and the $Ax$ is an $n \times 1$ vector too. so when $p=2$, it is the well known operator norm.