Ricky:

I think I see how to answer the problem under a stronger assumption. Rather than $\mathsf{AD}$, work in $$ {\sf AD}^+ + V=L({\mathcal P}({\mathbb R})). $$ This is a bit unsatisfying, since it is very possible the question can be answered assuming only $\mathsf{AD}$. In any case, $\mathsf{AD}^+$ is potentially harmless, since it may be that $\mathsf{AD}$ implies $\mathsf{AD}^+$; but it seems strange to need here the additional machinery that $\mathsf{AD}^+$ allows us. The assumption on the form of $V$ is more immediately troublesome, since it actually makes some (non-well-ordered) cardinals "invisible". (Meaning, there may be cardinals $\tau$ that inject into $2^{|{\mathbb R}|}$ but not in $L({\mathcal P}({\mathbb R}))$, so our assumption is potentially simplifying the question.)

Using $\mathsf{AD}^+ + V=L({\mathcal P}({\mathbb R}))$, Richard Ketchersid and I proved (in *A trichotomy theorem in natural models of ${\sf AD}^+$*, in **Set Theory and Its Applications**, Contemporary Mathematics, **533**, Amer. Math. Soc., Providence, RI, 2011, pp. 227-258) that *every* set either is well-orderable, or is at least as large as ${\mathbb R}$. (The paper is available here, and it also provides an introduction to $\mathsf{AD}^+$.)

Obviously, Lebesgue measure is not ${\mathbb R}$-additive, considering singletons, so this means it is enough to answer the question for well-ordered cardinals. But given a well-ordered union of sets of reals, we may assume they are pairwise disjoint (arguing inductively). And then it is standard that only countably many of them can be of positive measure. $(*)$

This shows (under the listed assumptions) that Lebesgue measure is $\kappa$-additive for all (well-orderable) $\kappa$ but not $\tau$-additive for *any* non-well-orderable cardinal $\tau$.

$(*)$ To see that the last paragraph follows, I need to argue that a well-ordered union of measure zero sets has measure zero. In fact, under $\mathsf{AD}$, a bit more holds, namely, $$ \iint_{[0,1]^2}f(x,y)dxdy=\iint_{[0,1]^2}f(x,y)dydx $$ for any bounded $f:[0,1]^2\to{\mathbb R}$ (in particular, the integrals are defined).

From this it is easy to conclude the claim about measure zero sets, e.g., by looking at minimal counterexamples and adapting the standard Sierpiński argument; note that this only uses $\mathsf{DC}({\mathbb R})$ (or even just $\mathsf{AC}_\omega(\mathbb R)$, which follows from $\mathsf{AD}$) and that all sets of reals are Lebesgue measurable.

(Unfortunately, although this Fubini theorem is not too hard, I am not sure of a reference for it. I remember vol. 5 of Fremlin's treatise has it as an exercise.) Let me add that this is a folklore result that is frequently used, as is the dual fact that a well-ordered union of meager sets is meager under $\mathsf{AD}$.