normalized elementwise L_1 norm is greater than spectral norm for correlation matrix?

Let $A=(A_{ij})$ be any $n \times n$ correlation matrix, $\|A\|_2$ be the spectral norm of $A$ (i.e., largest eigenvalue of $A$), and $|A|_1=\sum_{i=1}^n\sum_{j=1}^n|A_{ij}|$ be the elementwise $L_1$ norm of $A$.

Is that possible to find an absolute constant $C$ independent of $n$ such that

$$\|A\|_2\le Cn^{-1}|A|_1.$$

Note that $A$ is a correlation matrix, i.e., the diagonal entries of $A$ are $1$, and the off-diagonal entries are between $-1$ and $1$.

While the estimate is true with $C=1$ for the identity matrix and the all-one matrix, it is not true in general.
To see this, let $E_m$ denote the $m\times m$ all-one matrix, with $\|E_m\|_2=m$ and $|E_m|_1=m^2$.
Take the following matrix: $$A_{n,m} = \pmatrix{ E_m & 0 \\ 0 & I_n}.$$ Assume that the claimed estimate is true. That is, there exists $c>0$ such that $$\|A_{n,m}\|_2\le c (n+m)^{-1}|A_{n,m}|_1$$ for all $n,m$.
Since $\|A_{n,m}\|_2 = m$ and $|A|_1 = m^2 + n$, this is equivalent to $$m \le c \frac{m^2 + n}{m+n} \quad\forall n,m.$$ However, for $n\to\infty$ the right-hand side tends to $c$, and it follows $m\le c$ for all $m$, which is absurd.