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$\newcommand{\tnu}{\tilde\nu}$  Continuing Alex Ravsky's comment, we have \begin{equation*} 2^{2^n}\tnu_{n,1}=\sum_{j=0}^{2^n-1}S_j=S_0+(2^n-1)S_1, \tag{1} \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 each $j\ne0$ \begin{equation*} S_j=S_1=\sum_{k,l=0}^M\binom Mk \binom Ml |k-l|, \end{equation*} where \begin{equation*} M=M_n:=2^{n-1}. \tag{2} \end{equation*}

So, \begin{equation*} S_1=2^{2M} E|K-K'|, \tag{3} \end{equation*} where, for each natural $n$, $K=K_n$ and $K'=K'_n$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$), and hence $V_n:=(K-K')/\sqrt{M/4}$ converges in distribution to $Z-Z'$, where $Z$ and $Z'$ are independent standard normal r.v.'s.

Also, $EV_n^2=2<\infty$ for all $n$ and hence, by the de la Vallée-Poussin theorem, the $V_n$'s are uniformly integrable.

Therefore (see e.g. Theorem 3.5, p. 31), $E|V_n|\to E|Z-Z'|$ and hence, by (3), \begin{equation*} S_1\sim 2^{2M} \sqrt{\frac M4}\;E|Z-Z'|=2^{2M} \sqrt{\frac M4}\;\frac2{\sqrt\pi}. \tag{4} \end{equation*}

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. \tag{5} \end{equation*} Collecting the pieces (1), (5), (4), and (2), we finally get \begin{equation*} \tnu_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}. \end{equation*}

$\newcommand{\tnu}{\tilde\nu}$Continuing Alex Ravsky's comment, we have \begin{equation*} 2^{2^n}\tnu_{n,1}=\sum_{j=0}^{2^n-1}S_j=S_0+(2^n-1)S_1, \tag{1} \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 each $j\ne0$ \begin{equation*} S_j=S_1=\sum_{k,l=0}^M\binom Mk \binom Ml |k-l|, \end{equation*} where \begin{equation*} M=M_n:=2^{n-1}. \tag{2} \end{equation*}

So, \begin{equation*} S_1=2^{2M} E|K-K'|, \tag{3} \end{equation*} where, for each natural $n$, $K=K_n$ and $K'=K'_n$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$), and hence $V_n:=(K-K')/\sqrt{M/4}$ converges in distribution to $Z-Z'$, where $Z$ and $Z'$ are independent standard normal r.v.'s.

Also, $EV_n^2=2<\infty$ for all $n$ and hence, by the de la Vallée-Poussin theorem, the $V_n$'s are uniformly integrable.

Therefore (see e.g. Theorem 3.5, p. 31), $E|V_n|\to E|Z-Z'|$ and hence, by (3), \begin{equation*} S_1\sim 2^{2M} \sqrt{\frac M4}\;E|Z-Z'|=2^{2M} \sqrt{\frac M4}\;\frac2{\sqrt\pi}. \tag{4} \end{equation*}

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. \tag{5} \end{equation*} Collecting the pieces (1), (5), (4), and (2), we finally get \begin{equation*} \tnu_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}. \end{equation*}

$\newcommand{\tnu}{\tilde\nu}$\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=S_0+(2^n-1)S_1, \tag{1} \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 each $j\ne0$ \begin{equation*} S_j=S_1=\sum_{k,l=0}^M\binom Mk \binom Ml |k-l|, \end{equation*} where \begin{equation*} M=M_n:=2^{n-1}. \tag{2} \end{equation*}

So, \begin{equation*} S_1=2^{2M} E|K-K'|, \tag{3} \end{equation*} where, for each natural $n$, $K=K_n$ and $K'=K'_n$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$), and hence $V_n:=(K-K')/\sqrt{M/4}$ converges in distribution to $Z-Z'$, where $Z$ and $Z'$ are independent standard normal r.v.'s.

Also, $EV_n^2=2<\infty$ for all $n$ and hence, by the de la Vallée-Poussin theorem, the $V_n$'s are uniformly integrable.

Therefore (see e.g. Theorem 3.5, p. 31), $E|V_n|\to E|Z-Z'|$ and hence, by (3), \begin{equation*} S_1\sim 2^{2M} \sqrt{\frac M4}\;E|Z-Z'|=2^{2M} \sqrt{\frac M4}\;\frac2{\sqrt\pi}. \tag{4} \end{equation*}

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. \tag{5} \end{equation*} Collecting the pieces (1), (5), (4), and (2), we finally get \begin{equation*} \tnu_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}. \end{equation*}

$\newcommand{\tnu}{\tilde\nu}$ Continuing Alex Ravsky's comment, we have \begin{equation*} 2^{2^n}\tnu_{n,1}=\sum_{j=0}^{2^n-1}S_j=S_0+(2^n-1)S_1, \tag{1} \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 each $j\ne0$ \begin{equation*} S_j=S_1=\sum_{k,l=0}^M\binom Mk \binom Ml |k-l|, \end{equation*} where \begin{equation*} M=M_n:=2^{n-1}. \tag{2} \end{equation*}

So, \begin{equation*} S_1=2^{2M} E|K-K'|, \tag{3} \end{equation*} where, for each natural $n$, $K=K_n$ and $K'=K'_n$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$), and hence $V_n:=(K-K')/\sqrt{M/4}$ converges in distribution to $Z-Z'$, where $Z$ and $Z'$ are independent standard normal r.v.'s.

Also, $EV_n^2=2<\infty$ for all $n$ and hence, by the de la Vallée-Poussin theorem, the $V_n$'s are uniformly integrable.

Therefore (see e.g. Theorem 3.5, p. 31), $E|V_n|\to E|Z-Z'|$ and hence, by (3), \begin{equation*} S_1\sim 2^{2M} \sqrt{\frac M4}\;E|Z-Z'|=2^{2M} \sqrt{\frac M4}\;\frac2{\sqrt\pi}. \tag{4} \end{equation*}

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. \tag{5} \end{equation*} Collecting the pieces (1), (5), (4), and (2), we finally get \begin{equation*} \tnu_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}. \end{equation*}

$\newcommand{\tnu}{\tilde\nu}$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, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$). Also, $E|K-K'|^2=M/2$. So, by 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} where $Z,Z'$ are independent standard normal r.v.'s.
  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}

$\newcommand{\tnu}{\tilde\nu}$\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=S_0+(2^n-1)S_1, \tag{1} \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 each $j\ne0$ \begin{equation*} S_j=S_1=\sum_{k,l=0}^M\binom Mk \binom Ml |k-l|, \end{equation*} where \begin{equation*} M=M_n:=2^{n-1}. \tag{2} \end{equation*}

So, \begin{equation*} S_1=2^{2M} E|K-K'|, \tag{3} \end{equation*} where, for each natural $n$, $K=K_n$ and $K'=K'_n$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$), and hence $V_n:=(K-K')/\sqrt{M/4}$ converges in distribution to $Z-Z'$, where $Z$ and $Z'$ are independent standard normal r.v.'s.

Also, $EV_n^2=2<\infty$ for all $n$ and hence, by the de la Vallée-Poussin theorem, the $V_n$'s are uniformly integrable.

Therefore (see e.g. Theorem 3.5, p. 31), $E|V_n|\to E|Z-Z'|$ and hence, by (3), \begin{equation*} S_1\sim 2^{2M} \sqrt{\frac M4}\;E|Z-Z'|=2^{2M} \sqrt{\frac M4}\;\frac2{\sqrt\pi}. \tag{4} \end{equation*}

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. \tag{5} \end{equation*} Collecting the pieces (1), (5), (4), and (2), we finally get \begin{equation*} \tnu_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}. \end{equation*}