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 $\nu_{n,1}$ is much smaller than $2^{3n/2}$:
$$
\begin{array}{c|c|c|c|c}
n&2^n&\tilde \nu_{n,1}&\nu_{n,1}&\lfloor 2^{3n/2}\rfloor\\
\hline
0&1&1&1&1\\
1&2&1.5&2&2\\
2&4&4.25&6&8\\
3&8&11.65...&14&22\\
4&16&31.56...&40&64\\
5&32&??&\ge 96&181\\
6&64&??&??&512
\end{array}
$$

>**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 $\nu_{n,1}=o(2^{3n/2})$? 
>
>**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