Experimentally, a constant bound of $2/3$ should do, while the bounds above grow with n.

We can get a reasonable bound by using the Thue-Morse sequence to select $w_i$'s.  So we start with a weight of $+1$ for the largest $|g|$, and then weight smaller $|g|$'s with the inverse of the signs so far. 

\begin{equation}
\mathbb{E}\min_{w_i \in \left\{-1,1\right\}}|w_1g_1+\cdots+w_ng_n|
\ \le\ 
\mathbb{E}\left|\sum_{i=1}^n s_{n-i}\ |g|_{(i)}\right|
\end{equation}

where $|g|_{(i)}$ is the $i^{th}$ element after sorting the $|g|$'s, and $s_i$ is the $i^{th}$ element of [A106400][1].

E.g. if the $w$'s are 1.31, -0.25, 2.59, 0.68, -0.77, then this bound is |2.59 - 1.31 - 0.77 + 0.68 - 0.25|.

This gives an expectation of $(4-2\sqrt{2})/\sqrt{\pi}$ for $n=2$, using reasoning like Bjorn Kjos-Hanssen's.

Here is some Mathematica code for experimenting with 100 sets of $n$ random variables:

![code with histogram of signed sums for n=10000][3]

I got expectations for this bound around 0.18 with $n$ of 100,000 or 1,000,000.

$\\$

[Update:  We can use the same notation to prove that the expectation in the question is less than $E[\max|g_i|]$. Let $v_n = 1$, let $v_{j-1} = -\text{sign}( \sum_{i=j}^n v_i |g|_{(i)} )$, and then indeed $\left|\sum v_i |g|_{(i)} \right| < \max |g(i)|$.]


  [1]: https://oeis.org/A106400
  [2]: https://i.sstatic.net/2j71C.png
  [3]: https://i.sstatic.net/P1Bmu.png