TITLE: Regular $p$-norm of a matrix
QUESTION [9 upvotes]: Let $n \in \mathbb{N}$ and $p \in [1,\infty]$ be fixed and endow $\mathbb{C}^n$ with the $p$-norm $\|\cdot\|_p$. For every matrix $A \in \mathbb{C}^{n \times n}$ we denote the operator norm of $A$ as an operator on $\mathbb{C}^n$ by $\|A\|_p$, too. Moreover, let $|A|$ denote the matrix whose entries are the absolute values of the entries of $A$. 
The number $\| \,|A|\,\|_p$ is sometimes called the regular norm of $A$ (in particular in Banach lattice theory, where a norm on the space of regular operators is constructed this way).
We clearly have $\|A\|_p \le \| \,|A|\,\|_p$, and equality holds for $p = 1$ and $p = \infty$. For general $p$, Mark Meckes explained in his answer to this question that the Riesz-Thorin theorem implies the estimate
\begin{align*}
    \| \,|A|\,\|_p \le c_{n,p} \|A\|_p, \tag{E}
\end{align*}
where $c_{n,p} = n^{\frac{2}{p}(1 - \frac{1}{p})}$. 
Question: Is the estimate (E) sharp? If not, what is the optimal constant for the estimate (depending on $n$ and $p$)?
Remarks:

Clearly, the inequality (E) is sharp (and actually an equality) for $p = 1$ and $p = \infty$.
Let $U \in \mathbb{C}^{n\times n}$ be such that $\frac{1}{\sqrt{n}}U$ is a unitary matrix and such that each entry of $U$ has modulus $1$ (such a matrix exists for every $n$; for instance, we can choose $U$ to be the transformation matrix of the Fourier transform on $\mathbb{Z}/n\mathbb{Z}$). Then one can prove by employing Hölder's inequality several times that $\|\, |U| \, \|_p = n$ for each $p \in [1,\infty]$, and that $\|U\|_p = n^{1/p}$ for $p \in [1,2]$, while $\|U\|_p = n^{1-1/p}$ for $p \in [2,\infty]$. Thus, we obtain $\|\,|U|\,\|_p = \tilde c_{n,p} \|U\|_p$, where
\begin{align*}
  \tilde c_{n,p} = 
  \begin{cases}
    n^{1-1/p} \quad & \text{for } p \in [1,2], \\
    n^{1/p} \quad & \text{for } p \in [2,\infty].
  \end{cases}
\end{align*} 
We have $\tilde c_{n,p} \le c_{n,p}$, and the existence of $U$ shows that the optimal constant for (E) is somewhere in between. 
Note that $\tilde c_{n,2} = c_{n,2} = n^{1/2}$, so we know that the estimate (E) is sharp for $p=2$; this is also mentioned in Mark Meckes' answer quoted above.

Note: A loosely related question concerned with the Schatten $p$-norm of a matrix can be found here.
Edits made 2018-06-03: 

I've replaced the real scalar field with $\mathbb{C}$ since this seems to be more appropriate for the question. 
I've added the example matrix $U$ to show that the number $\tilde c_{n,p}$ is a lower bound for the optimal constant.

REPLY [6 votes]: It is trivially $n^{1/p}$  for $p>2$ (and, therefore, $n^{1-\frac 1p}$ for $1<p<2$ because the norm of the adjoint operator in the dual space is the same as the norm of the operator in the space itself). To see it, choose the vector $x$ of $\ell^p$ norm $1$ so that
$$
\sum_i\left|\sum_j |a_{ij}|x_j\right|^p=\||A|\|^p
$$
Now choose $i$ so that 
$$
\left|\sum_j |a_{ij}|x_j\right|^p\ge \frac 1n\||A|\|^p
$$
Rotate $x_j$ to $y_j$ preserving the absolute values so that for this particular $i$, we have 
$$
\sum_j |a_{ij}|x_j=\sum_j a_{ij}y_j
$$
Then 
$$
\|A\|^p\ge \sum_i\left|\sum_j a_{ij}y_j\right|^p\ge \frac 1n\||A|\|^p
$$
and we are done.
What "Riesz-Thorin" are we talking about? Much less trivial questions get closed in no time and this one got 8 upvotes and 2 favorites...