$\newcommand{\tnu}{\tilde\nu}$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}