What is the expected root mean square determinant of an $n\times n$ matrix? The expected mean determinant of random $n\times n$ matrices of $0$'s and $1$'s is $0$.
What is the expected root mean square determinant? e.g. $\frac{\sqrt{3}}{2\sqrt{2}}$ for a $2\times 2$
 A: As you say, the expected mean determinant is zero:
$$
\def\expect#1{\left\langle#1\right\rangle}
\begin{eqnarray}
\expect{\det A}
&=&
\expect{\sum_\pi\sigma(\pi)\prod_iA_{i\pi(i)}}
\\
&=&
\sum_\pi\sigma(\pi)\expect{\prod_iA_{i\pi(i)}}
\\
&=&
\sum_\pi\sigma(\pi)\expect{\prod_iA_{ii}}
\\
&=&
\expect{\prod_iA_{ii}}\sum_\pi\sigma(\pi)
\\
&=&
0\;.
\end{eqnarray}
$$
The expected mean squared determinant of an $n\times n$ matrix can be calculated similarly:
$$
\begin{eqnarray}
\expect{(\det A)^2}
&=&
\expect{\left(\sum_\pi\sigma(\pi)\prod_iA_{i\pi(i)}\right)^2}
\\
&=&
\expect{\left(\sum_\pi\sigma(\pi)\prod_iA_{i\pi(i)}\right)\left(\sum_\rho\sigma(\rho)\prod_iA_{i\rho(i)}\right)}
\\
&=&
\expect{\sum_{\pi,\rho}\sigma(\pi)\sigma(\rho)\prod_iA_{i\pi(i)}\prod_iA_{i\rho(i)}}
\\
&=&
\expect{\sum_{\pi,\rho}\sigma(\pi^{-1}\rho)\prod_iA_{ii}\prod_iA_{i(\pi^{-1}\rho)(i)}}
\\
&=&
n!\expect{\sum_{\pi}\sigma(\pi)\prod_iA_{ii}\prod_iA_{i\pi(i)}}\;.
\\
&=&
n!\sum_{\pi}\sigma(\pi)\expect{\prod_iA_{ii}\prod_iA_{i\pi(i)}}\;.
\end{eqnarray}
$$
The expectation value in the last line has $n$ factors of $1/2$ from the diagonal elements and another factor of $1/2$ for each $i$ that is not a fixed point of $\pi$. Thus we have
$$
\begin{eqnarray}
\expect{(\det A)^2}
&=&
\frac{n!}{2^n}\sum_\pi\sigma(\pi)M_{i\pi(i)}
\\
&=&
\frac{n!}{2^n}\det M
\end{eqnarray}
$$
with
$$M_{ij}=\begin{cases}\vphantom{\frac12}1&i=j\;,\\\frac12&i\ne j\;.\end{cases}$$
This matrix has $n-1$ eigenvalues of $1/2$ and one eigenvalue of $(n+1)/2$, and so 
$$
\begin{eqnarray}
\expect{(\det A)^2}
&=&
\frac{n!}{2^n}\frac{n+1}{2^n}
\\
&=&
\frac{(n+1)!}{2^{2n}}
\end{eqnarray}
$$
and
$$\sqrt{\expect{(\det A)^2}}=\frac{\sqrt{(n+1)!}}{2^n}\;,$$
in agreement with your result for $n=2$.
See also OEIS sequence A055137 for more on the connection to fixed points of permutations.
The calculation doesn't rely on the distribution of the matrix elements up to the point where we plug in the factors of $1/2$; evaluating the result for matrix elements uniformly chosen as $\pm1$ instead and noting that in this case any unpaired matrix element causes the expectation value to vanish, we obtain a mean squared determinant of $n!$ for a random matrix of $\pm1$s.
P.S.: Another way to evaluate
$$
\expect{(\det A)^2}=n!\sum_{\pi}\sigma(\pi)\expect{\prod_iA_{ii}\prod_iA_{i\pi(i)}}
$$
just occurred to me. There are $\binom nk$ ways to choose $n-k$ fixed points for a permutation. The rest of the permutation is a derangement of $k$ objects, and in the inclusion-exclusion sum for the number of derangements,
$$
k! \sum_{i=0}^k \frac{(-1)^i}{i!}
$$
(see the Wikipedia article), all terms except the last two count the same number of even and odd permutations, whereas the last two only count even permutations, so the excess of even over odd permutations in this sum is
$$
(-1)^k+(-1)^{k-1}k\;.
$$
Thus
\begin{align}
\expect{(\det A)^2}&=n!\sum_{\pi}\sigma(\pi)\expect{\prod_iA_{ii}\prod_iA_{i\pi(i)}}\\
&=\frac{n!}{2^n}\sum_{k=0}^n\binom nk\left((-1)^k+(-1)^{k-1}k\right)\left(\frac12\right)^k\\
&=\frac{n!}{2^n}\left(\left(\frac12\right)^n+n\left(\frac12\right)^n\right)\\
&=\frac{(n+1)!}{2^{2n}}\;,
\end{align}
in agreement with the result obtained using the determinant.
