You can relate $B_2$ balls to the domain of analyticity of your function $f$. For instance, if $d=1$, $f\in B_2(r,L)$ implies that $f$ is analytic on $\{z : \vert z\vert < \sqrt{e^r}\}$ and reciprocally, if $f$ is analytic on $\{z : \vert z\vert < \sqrt{e^\rho}\}$ then $f\in B_2(r,L)$ for all $r<\rho$.

To put it differently, if you can say something about the domain of analyticity of your function $f$ you can then use Cauchy's formula to control the decay of the coefficients $a_k$, and therefore you could directly get an answer to your Question 3.

You should have similar relations in higher dimension. Indeed, the only thing you really needs is Cauchy integral formula, which can be generalized in higher dimension (see e.g. Hörmander's book "An Introduction to Complex Analysis in Several Variables"). From that formula, if you assume that $f$ is analytic on $\mathcal{B}_\rho := \{ z\in \mathbb{C}^d : \vert z_i \vert < \rho_i,\ i=1,\ldots,d \}$, then for any $s=(s_1,\ldots,s_d)$ s.t. $0<s_i<\rho_i$ for all $i$, you get something like
$$
\vert a_k \vert = \left\vert \frac{\partial_k f (0)}{k!} \right\vert \leq \frac{\sup_{\mathcal{B}}\vert f\vert}{(2\pi)^d} \frac{1}{s^k}.
$$
As soon as you can take each $s_i>1$, then you get a control on the decay of your coefficients, and you can say that $f$ belongs to some of your $B_2$ balls, or directly try and answer your question 3.