The question makes the most sense if we assume $V=L$, since otherwise it could be that every set of reals in $L$ is countable and hence measurable.
If $V=L$, the answer is $\omega_1+1$. Every set in $L_{\omega_1^L}$ is countable in $L$ and hence measurable, and so it cannot be $\omega_1$ or less. But at stage $\omega_1$, the set $2^\omega$ is definable in $L_{\omega_1}$, and we can also define the $L$-order on these reals. So we can define a Vitali set at this level. Namely, we can define the tail-equivalence relation on binary sequences, eventual agreement, and I can pick the $L$-least member of each equivalence class, all definably over $L_{\omega_1^L}$. This set is not measurable using the coin-flipping probability measure on $2^\omega$ in $L$, since we can consider the sets that arise by flipping finitely many digits. These are disjoint, for all the countably many ways to flip digits differently, but the resulting sets all have the same measure as the original, and they cover $2^\omega$, contradiction, just as in the usual Vitali argument.
So the nonmeasurable sets show up at the same time as the whole set $2^\omega$ itself, namely, in $L_{\omega_1+1}$.
It seems that this argument works provided that $(2^\omega)^L$ has positive measure, since it will be covered by countably infinitely many disjoint copies of a fixed set, all with the same measure. And if $(2^\omega)^L$ does not have positive measure, then either it itself is nonmeasurable and exists at this level, or it is measure zero, in which case there will be no nonmeasurable sets in $L$. So the answer is $\omega_1^L+1$, without assuming $V=L$, provided that there is indeed some nonmeasurable set in $L$.