Given any $n\in\mathbb N$, consider the [the Sylvester-Hadamard-Walsh matrix][1] $M=(a_{i,j})_{i,j\in 2^n}$ 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):
$$
\begin{array}{c|c|c|c|c|c|c}
n&2^n&\tilde \nu_{n,1}&\nu_{n,1}&\lfloor 2^{3n/2}\rfloor&2^{3n/2}/\tilde\nu_{n,1}\\
\hline
0&1&1&1&1&1\\
1&2&1.5&2&2&1.33...\\
2&4&4.25&6&8&0.94...\\
3&8&11.65...&14&22&0.68...\\
4&16&31.56...&40&64&0.50...\\
5&32&85.41...&\ge 96&181&0.37...\\
6&64&232.28..&??&512&0.27...\\
7&128&636.09...&??&1448&0.20...\\
8&256&1754.09...&??&4096&0.14...\\
9&512&4866.56...&??&11585&0.10...\\
\end{array}
$$

>**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 $1<\lambda_1<\frac32$?


  [1]: https://en.wikipedia.org/wiki/Walsh_matrix