Given any $n\in\mathbb N$, consider the [the Sylvester-Hadamard-Walsh matrix](https://en.wikipedia.org/wiki/Walsh_matrix) $M=(a\_{i,j})\_{i,j\in 2^n}$ of size $2^n\times 2^n$ and for a number $p\in[1,\infty)$, let
$$\nu\_{n,p}=\max\_{F\subseteq 2^n}\Big(\sum\_{j\in 2^n}\big|\sum\_{i\in F}a\_{i,j}\big|^p\Big)^{1/p}\quad\mbox{and}\quad \tilde \nu\_{n,p}=\frac1{2^{2^n}}\sum\_{F\subseteq 2^n}\Big(\sum\_{j\in 2^n}\big|\sum\_{i\in F}a\_{i,j}\big|^p\Big)^{1/p}.$$
For $p=2$, the Pithagoras Theorem and the orthogonality of the rows of the matrix $M$ imply that $\nu\_{n,2}=2^n$. Using this equality, it is easy to show that $\nu\_{n,p}=2^n$ for all $p\in[2,\infty)$.


If $p\in[1,2]$, then by the Holder inequality, we obtain
$$2^n\le\nu\_{n,p}\le 2^{n(\frac1p+\frac12)}.$$ In particular, $2^n\le\nu\_{n,1}\le 2^{3n/2}$. On the other hand, computer calculations show that $\tilde \nu\_{n,1}$ and $\nu\_{n,1}$ are much smaller than $2^{3n/2}$ (the values of $\tilde\nu\_{n,1}$ are calculated using the formula
$$\tilde\nu\_{n,1}=\frac1{2^{2^n}}\Big(\sum\_{i=0}^{2^n}{2^n\choose i}i+2(2^n-1)\sum\_{0\le i<j\le 2^n}{2^n\choose i}{2^n\choose j}(j-i)\Big)$$
of Alex Ravsky suggested in his comment):
$$
\begin{array}{c|c|c|c|c|c|c}
n&2^n&\tilde \nu\_{n,1}&\nu\_{n,1}&\lfloor 2^{3n/2}\rfloor&\tilde\nu\_{n,1}/2^{3n/2}\\
\hline
0&1&1&1&1&1\\
1&2&1.5&2&2&0.53...\\
2&4&4.25&6&8&0.53...\\
3&8&11.65...&14&22&0.51...\\
4&16&31.56...&40&64&0.49...\\
5&32&85.41...&\ge 96&181&0.47...\\
6&64&232.28..&??&512&0.45...\\
7&128&636.09...&??&1448&0.43...\\
8&256&1754.09...&??&4096&0.42...\\
9&512&4866.56...&??&11585&0.42...\\
\end{array}
$$



> 
> **Problem 1.** Is $\nu\_{n,1}\ge\frac12 2^{3n/2}$? Is $\tilde\nu\_{n,1}\ge \varepsilon 2^{3n/2}$ for some $\varepsilon>0$?
> 
> 
> **Problem 2.** Is $\tilde\nu\_{n,1}=o(2^{3n/2})$? Is $\nu\_{n,1}=o(2^{3n/2})$?
> 
> 
> **Problem 3.** Find nontrivial lower and upper bounds on the number $$\lambda\_1=\limsup\_{n\to\infty}\frac1n\log\_2(\nu\_{n,1}).$$ Is $1<\lambda\_1<\frac32$?
> 
> 
> 



The largest $\ell_p$-norm of a sum of rows of a Sylvester-Hadamard-Walsh matrix
$\newcommand{\tnu}{\tilde\nu}$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](https://en.wikipedia.org/wiki/Uniform_integrability#Relevant_theorems), the $V\_n$'s are uniformly integrable.


Therefore (see e.g. [Theorem 3.5, p. 31](http://cermics.enpc.fr/%7Emonneau/Billingsley-2eme-edition.pdf)), $E|V\_n|\to E|Z-Z'|$ 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\*}


