I note that the ordinals of L are the same as V, so I guess that it has no $\Pi_1^1$ consequences. On the other hand Wikipedia tells me that it asserts the existance of a $\Delta_2^1$ non-measurable set of reals. "Measurable" involves third-order concepts but I know that there's often a "coding trick" that gets around that sort of thing, so I guess it has some analytic consequences.
Of course I am guessing -- I am not very good at this stuff. But I am curious. What's the least point in the analytic hierarchy where V=L matters (if any)?
Shoenfield's absoluteness theorem ensures that any $\Sigma^1_2$ statement (and therefore, any $\Pi^1_2$ one) has the same truth value in $L$ as in $V$.
On the other hand, $\Sigma^1_3$ statements differ in general whether they are in $V$ or in $L$. For a silly example, $\exists x\in{\mathbb R}\,(x\notin L)$ can be rewritten as a $\Sigma^1_3$ statement, which is false in $L$ but true in general. Given a $\Sigma^1_2$ formula $\phi(x,y)$, the statement ``$\phi$ defines a well-ordering of the reals'' is $\Pi^1_3$. Again, there is a specific such $\phi$ for which this is true in $L$ and false in general.
("In general" means here that it can be made false by forcing. There are deeper discrepancies if one allows large cardinals into the picture.)