Let us work in terms of Cohen's $d$ and then convert to $g$.
It is known that the variance of $d$ is
$$
V_d = \frac{n_1 + n_2}{n_1n_2} + \frac{d^2}{2(n_1 + n_2)}
$$
where $n_1$ and $n_2$ are the sample sizes per group.
Suppose we in fact have $g$, we know that
$$
g = J d
$$
where
$$
J = 1 - \frac{3}{4\nu - 1}
$$
where $\nu$ is the df, that is $n_1 + n_2 - 2$
and
$$
V_g = J^2 V_d
$$
So by backcalculation from $g$ to $d$, computing the variance there and then converting back we get the sampling variance for $g$. The required standard error is then the square root. If only the overall sample size is known then setting $n_1 = n_2 = n/2$ would be defensible.