Consider a random circulant $n\times n$ $0$-$1$ matrix $A$. Let the probability $P(A_{1,i} = 1) = 1/\sqrt{n}$ and all the elements of the first row be probabilistically independent.
We know that the expected value of the diagonal elements of $S=AA^T$ is $\sqrt{n}$ and that the off-diagonal have mean $1$.
If $n$ is large, what can we say about the eigenvalues of $S$?
In particular, numerically one can see that the mean eigenvalue is $\sqrt{n}$? How can one prove this?
The mean eigenvalue follows pretty easily from the work you've done. As you mentioned, the diagonal elements of $S$ are identical and equal to the sum of 1's in the vector that defines the circulant matrix
$$S_{ii} \sim \operatorname{Binomial}(n,1/\sqrt{n}) $$
From the linearity of the expectation, the expected mean eigenvalue is:
$$E(\lambda(S))=E\left(\frac{\operatorname{tr} (S)}{n}\right)=E(S_{ii})=\sqrt{n}$$