= Solution
Start at $x=(x_1,\ldots,x_n)$ with all $x_i>0$, and let $\tau_i$ be the first zero of the $i$th coordinate. The orthant exit time is $T=\min_i\tau_i$. The coordinate processes are <independent>, so
$$
\mathbb P_x(T>t)=\prod_{i=1}^n\mathbb P_{x_i}(\tau_i>t)
=\prod_{i=1}^n\left[2\Phi(x_i/\sqrt t)-1\right].
$$
Here the <Brownian reflection principle> gives $\mathbb P_a(\tau\leq t)=2\mathbb P(B_t\leq-a)$; $\Phi$ is the <standard normal distribution> function. Since $\Phi(z)-1/2=z/\sqrt{2\pi}+o(z)$ as $z\to0$,
$$
\mathbb P_x(T>t)\sim(2/\pi)^{n/2}(x_1\cdots x_n)t^{-n/2}\quad(t\to\infty).
$$
The <tail integral formula for expectation>, $\mathbb E_xT=\int_0^\infty\mathbb P_x(T>t)dt$, has a bounded integrand on $(0,1)$ and converges at infinity exactly when $n/2>1$. Therefore the <Brownian exit-time expectation in an orthant> gives
$$
\boxed{g_{D_1}=\infty,\qquad g_{D_2}=\infty,\qquad g_{D_3}<\infty\text{ everywhere in }D_3.}
$$
The divergence for $n=2$ is logarithmic; almost-sure finiteness of the exit time does not imply a finite <expectation>.
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