I have not seen this question discussed in the literature either, but I will propose an answer here.
To begin with, I think that since $n^{n/2}$ is the maximum possible absolute value that the determinant of any $\{-1,1\}$-matrix can have (Hadamard upper bound), the constant $c$ in the question (if it exists) must necessarily be $\geq 2$.
I will now argue that any $c>2$ has the property described in the question. I will do that by sketching a derivation of the following lower bound for the largest $|\det A|$, which is stronger asymptotically for any $c>2$ than the bound $n^{n/c}$ suggested in the question.
Proposition:
There exists a real constant $\alpha > 0$, so that for any positive integer $n \geq 3$ there is at least one circulant matrix $A \in \{-1,1\}^{n \times n}$ that obeys $|\det A| > (\alpha n)^{n/2}$. Moreover, $\alpha = \sqrt2/{\rm e} \approx 0.52$ (and thus also any smaller $\alpha$) has this property.
The above proposition can be proved as follows. Using the same technique as described in this previous answer, it can be shown that the mean value of $|\det A|^2$ over all $n \times n$ circulant $\{-1,1\}$-matrices $A$ is bounded from below by $2^{\lfloor \frac{n-1}2 \rfloor} n!$, for any $n$ except $n=2$, where that mean value is zero. (In the previous answer, a lower bound $2^{-2n+\lfloor \frac{n-1}2 \rfloor} (n+1)!$ was derived for circulant $\{0,1\}$-matrices instead.)
Using Stirling's lower bound on $n!$ we then obtain the following sequence of lower bounds on the mean value of $|\det A|^2$:
$$
2^{\lfloor \frac{n-1}2 \rfloor} n! > 2^{\frac{n}2-1} \sqrt{2\pi n}\ \left(\frac{n}{\rm e}\right)^n > \left(\frac{\sqrt2}{\rm e} n \right)^n \ .
$$
The last of these bounds implies that among these matrices $A$ there must be at least one with $|\det A| > (\alpha n)^{n/2}$, where $\alpha =\sqrt2/{\rm e}$, which concludes the proof.
