$\def\Tr{\mathrm{Tr}}\def\Mat{\mathrm{Mat}}$I've been thinking about this problem a bunch, and I think the correct bound  is
$$ \sum_{i,j} |A_{ij}| \leq \left( \cot \frac{\pi}{2n}  \right)|A|_{(1)}. $$
As $n \to \infty$, we have $\cot \tfrac{\pi}{2n} \sim \tfrac{2n}{\pi}$, so this matches the $\pi$ bound that fedja proved for $k=2$.
Unfortunately, I can't prove much, but here are ideas that might help someone else.

I'll write $\sigma_1(A) \geq \sigma_2(A) \geq \cdots$ for the singular values of $A$. Note that we have
$$\sum |A_{ij}| = \max_{P \in \mathrm{Mat}_n(\pm 1)} \Tr(AP) $$
and 
$$|A|_{(1)} = \max_{Q \in O(n)} \Tr(AQ). $$
Here $P$ is ranging over $\pm 1$ matrices, and $Q$ is ranging over the orthogonal group.

We may replace the orthogonal group by its convex hull without changing the max. The convex hull of $O(n)$ is the set of matrices of operator norm $\leq 1$; call that $B_1$.
So
$$|A|_{(1)} = \max_{R \in B_1} \Tr(AR). $$

As a warm up, let's consider the best inequality we can prove of the form $\sum |A_{ij}| \leq C |A|_{(1)}$ without imposing that the diagonal is $0$. The answer is that the best is $C = n$, and that is easy to prove by elementary means, but I want to demonstrate my approach instead. So we want to find a $C$ such that, for every $\pm 1$ matrix $P$ and for every matrix $A$, we have $\Tr(AP) \leq C \max_{R \in B_1} \Tr(AR)$. Since $B_1$ is convex, this is the same as asking for $C$ such that $P \in C B_1$. In other words, we want to bound $\sigma_1(P)$ for $P$ in $\Mat_n(\pm 1)$. It wouldn't be hard to obtain the bound $n$ from here, but we move on.

Let's leave the warm up and get to the real problem. What we actually want is that $\Tr(AP) \leq C \max_{R \in B_1} \Tr(AR)$ for $A$ having zero diagonal. Thus, we only need $\pi(P)$ to lie in $\pi(C B_1)$, where $\pi$ is orthogonal projection onto matrices of diagonal $0$. In other words, we want $P$ to lie in $CB_1 + D$ where $D$ is the vector space of diagonal matrices. So we come to the following problem:

<b>Problem 1:</b> Find the best constant $C_1$ such that, for every $\pm 1$ matrix $P$, there is a diagonal matrix $D$ with $\sigma_1(P+D) \leq C_1$. 

Unfortunately, it seems hard to even guess a rule for choosing the optimal $D$. For example, if $P$ is identically $1$, the best choice of $D$ is $-\frac{n}{2} \mathrm{Id}_n$.

Having no success here, I move on to the case of $A$ skew symmetric. We now can consider only skew symmetric $P$ (which are $0$ on the diagonal and $\pm 1$ off the diagonal.) For such a $P$, we now want to solve the problem: 

<b>Problem 2:</b> Find the best constant $C_1$ such that, for every skew-symmetric $\pm 1$ matrix $P$, there is a symmetric matrix $H$ with $\sigma_1(P+H) \leq C_1$. 

Fortunately, here I can make progress. It turns out that the symmetric matrix is irrelevant!

<b>Lemma:</b> Let $P$ be a skew symmetric matrix and $H$ a symmetric matrix. Then $\sigma_1(P+H) \geq \sigma_1(P)$.

<b>Proof:</b> Since $P$ is skew symmetric, it is diagonalizable over $\mathbb{C}$ with purely imaginary eigenvalues, and the largest such is $i \sigma_1(P)$. Let $v$ be an eigenvector with $P v = i \sigma_1 v$. Writing $\dagger$ for the conjugate transpose, normalize $v^{\dagger} v =1$. Then $\sigma_1(P+H) \geq | v^{\dagger} (P+H) v | = |i \sigma_1 + v^{\dagger} H v|$. But $v^{\dagger} H v$ is real, so $|i \sigma_1 + v^{\dagger} H v| \geq \sigma_1$. $\square$.

Thus, we have reduced to the problem:

<b>Problem 3:</b> Find the largest operator norm of any skew-symmetric $\pm 1$ matrix. 

I have checked for $n \leq 6$, and the largest operator norm is always achieved by the matrix which is $1$'s above the diagonal and $-1$'s below it. (As well as by the many other matrices which are conjugate to this one by signed permutation matrices.) This matrix can be explicitly diagonalized:
The eigenvectors are of the form $(1, \zeta, \zeta^2, \ldots, \zeta^{n-1})$ where $\zeta = \exp(\pi i (2j+1)/(2n))$. The corresponding eigenvalues are $i \cot \tfrac{(2j+1) \pi}{2n}$. In particular, the largest singular value is $\cot \tfrac{\pi}{2n}$, thus explaining my guess.

I am guessing this is optimal for Problem 1 as well as Problem 2, but this is based on a very weak intuition that skew symmetric choices are good, plus fedja's answer.