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.
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:
I got expectations for this bound around 0.18 with $n$ of 100,000 or 1,000,000.
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[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 $\left|\sum v_i |g|_{(i)} \right| < \max |g(i)|$.]