For any numbers $g_1,\dots,g_n$, we have $$ \min_{w_i=\pm 1} \left|\sum_i w_ig_i\right|\le \max_i |g_i|$$ since we can let $w_i=\mathrm{sign}(g_i)$ when $\sum_{j\le i} |g_j|<\sum_{j> i} |g_j|$ and $w_i=-\mathrm{sign}(g_i)$ otherwise. Then $$\mathbb E\max_{i=1}^n |g_i|=\int_0^\infty \Pr\left[\max_i |g_i|)\ge x\right] dx$$ $$=\int_0^\infty 1-\Pr\left[\max_i |g_i|\le x\right]dx=\int_0^\infty 1-\left(\Pr\left[|g_1|\le x\right]\right)^ndx$$ $$=\int_0^\infty 1-(\Phi(x)-\Phi(-x))^ndx,\qquad\Phi=\text{standard normal cdf}$$ $$=\int_0^\infty 1-(1-2\Phi(-x))^ndx.$$ I suppose one can say more at this point but I'll just note that Wolfram Alpha has some further info about such integrals. For $n=1$ it's $\int 2\Phi(-x))dx=\frac{2}{\sqrt{2\pi}}=\sqrt{2/\pi}$ as it should be, and for $n=2$ it is $$ \int_0^\infty 4\Phi(-x)-4\Phi(-x)^2\,dx=4 \left(\frac1{\sqrt{2\pi}} - \frac{\sqrt 2-1}{\sqrt{2\pi}}\right) = 4 \frac{2-\sqrt 2}{\sqrt{2\pi}} = 4\frac{\sqrt 2-1}{\sqrt\pi} $$ which is already better than @user35593's estimate.

For any numbers $g_1,\dots,g_n$, we have $$ \min_{w_i=\pm 1} \left|\sum_i w_ig_i\right|\le \max_i |g_i|$$ since you can separate the $g_i$ into two piles whose sum of absolute values are about equally large (see @MattF.'s answer for a precise formulation). Then $$\mathbb E\max_{i=1}^n |g_i|=\int_0^\infty \Pr\left[\max_i |g_i|)\ge x\right] dx$$ $$=\int_0^\infty 1-\Pr\left[\max_i |g_i|\le x\right]dx=\int_0^\infty 1-\left(\Pr\left[|g_1|\le x\right]\right)^ndx$$ $$=\int_0^\infty 1-(\Phi(x)-\Phi(-x))^ndx,\qquad\Phi=\text{standard normal cdf}$$ $$=\int_0^\infty 1-(1-2\Phi(-x))^ndx.$$ I suppose one can say more at this point but I'll just note that Wolfram Alpha has some further info about such integrals. For $n=1$ it's $\int 2\Phi(-x))dx=\frac{2}{\sqrt{2\pi}}=\sqrt{2/\pi}$ as it should be, and for $n=2$ it is $$ \int_0^\infty 4\Phi(-x)-4\Phi(-x)^2\,dx=4 \left(\frac1{\sqrt{2\pi}} - \frac{\sqrt 2-1}{\sqrt{2\pi}}\right) = 4 \frac{2-\sqrt 2}{\sqrt{2\pi}} = 4\frac{\sqrt 2-1}{\sqrt\pi} $$ which is already better than @user35593's estimate.

For any numbers $g_1,\dots,g_n$, we have $$ \min_{w_i=\pm 1} \left|\sum_i w_ig_i\right|\le \max_i |g_i|$$ since we can let $w_i=\mathrm{sign}(g_i)$ when $\sum_{j<i} |g_j|<\sum_{j\ge i} |g_j|$ and $w_i=-\mathrm{sign}(g_i)$ otherwise. Then $$\mathbb E\max_{i=1}^n |g_i|=\int_0^\infty \Pr\left[\max_i |g_i|)\ge x\right] dx$$ $$=\int_0^\infty 1-\Pr\left[\max_i |g_i|\le x\right]dx=\int_0^\infty 1-\left(\Pr\left[|g_1|\le x\right]\right)^ndx$$ $$=\int_0^\infty 1-(\Phi(x)-\Phi(-x))^ndx,\qquad\Phi=\text{standard normal cdf}$$ $$=\int_0^\infty 1-(1-2\Phi(-x))^ndx.$$ I suppose one can say more at this point but I'll just note that Wolfram Alpha has some further info about such integrals. For $n=1$ it's $\int 2\Phi(-x))dx=\frac{2}{\sqrt{2\pi}}=\sqrt{2/\pi}$ as it should be, and for $n=2$ it is $$ \int_0^\infty 4\Phi(-x)-4\Phi(-x)^2\,dx=4 \left(\frac1{\sqrt{2\pi}} - \frac{\sqrt 2-1}{\sqrt{2\pi}}\right) = 4 \frac{2-\sqrt 2}{\sqrt{2\pi}} = 4\frac{\sqrt 2-1}{\sqrt\pi} $$ which is already better than @user35593's estimate.

For any numbers $g_1,\dots,g_n$, we have $$ \min_{w_i=\pm 1} \left|\sum_i w_ig_i\right|\le \max_i |g_i|$$ since we can let $w_i=\mathrm{sign}(g_i)$ when $\sum_{j\le i} |g_j|<\sum_{j> i} |g_j|$ and $w_i=-\mathrm{sign}(g_i)$ otherwise. Then $$\mathbb E\max_{i=1}^n |g_i|=\int_0^\infty \Pr\left[\max_i |g_i|)\ge x\right] dx$$ $$=\int_0^\infty 1-\Pr\left[\max_i |g_i|\le x\right]dx=\int_0^\infty 1-\left(\Pr\left[|g_1|\le x\right]\right)^ndx$$ $$=\int_0^\infty 1-(\Phi(x)-\Phi(-x))^ndx,\qquad\Phi=\text{standard normal cdf}$$ $$=\int_0^\infty 1-(1-2\Phi(-x))^ndx.$$ I suppose one can say more at this point but I'll just note that Wolfram Alpha has some further info about such integrals. For $n=1$ it's $\int 2\Phi(-x))dx=\frac{2}{\sqrt{2\pi}}=\sqrt{2/\pi}$ as it should be, and for $n=2$ it is $$ \int_0^\infty 4\Phi(-x)-4\Phi(-x)^2\,dx=4 \left(\frac1{\sqrt{2\pi}} - \frac{\sqrt 2-1}{\sqrt{2\pi}}\right) = 4 \frac{2-\sqrt 2}{\sqrt{2\pi}} = 4\frac{\sqrt 2-1}{\sqrt\pi} $$ which is already better than @user35593's estimate.