$\newcommand{\tnu}{\tilde\nu}$Continuing Alex Ravsky's comment, we have 
\begin{equation}
	2^{2^n}\tnu_{n,1}=\sum_{j=0}^{2^n-1}S_j,
\end{equation}
where 
\begin{equation}
	S_j:=\sum_{F\subseteq[2^n]}\Big|\sum_{i\in F}a_{i,j}\Big|,
\end{equation}
and for $j\ne0$
\begin{equation}
	S_j=\sum_{k,l=0}^M\binom Mk \binom Ml |k-l|,
\end{equation}
where $M:=2^{n-1}$. 

So, for $j\ne0$, 
\begin{equation}
	S_j=2^{2M} E|K-K'|,
\end{equation}
where $K,K'$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem and uniform integrability, for $j\ne0$, 
\begin{equation}
	S_j\sim 2^{2M}  \sqrt{\frac M4}\;E|Z-Z'|=2^{2M}  \sqrt{\frac M4}\;\frac2{\sqrt\pi}
\end{equation}
(as $n\to\infty$), where $Z,Z'$ are independent standard normal r.v. 
Also,
\begin{equation}
	S_0=\sum_{F\subseteq[2^n]}\sum_{i\in F}1=\sum_{k=0}^{2M}\binom{2M}k k=2^{2M}M.
\end{equation}
Collecting the pieces, we get 
\begin{equation}
	\tnu_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}.
\end{equation}