It depends what you mean by "logic". Is second-order logic a logic? If so, then the answer to your question is no: second-order logic has no associated sequent calculus, since it is not compact (so in particular there is no way to represent entailment in second-order logic in a finitary way).
Note that there are other non-compact logics - say, infinitary logic $\mathcal{L}_{\omega_1\omega}$. However, second-order logic is a more compelling counterexample: unlike $\mathcal{L}_{\omega_1\omega}$, which does have a kind of proof system associated to it (via an infinitary sequent calculus, developed by Lopez-Escobar and later Barwise, if I recall correctly) which is reasonably set-theoretically absolute, second-order logic is just completely terrible.
Specifically, the question of validity in second-order logic - what second-order sentences are true in all models - is a fundamentally set-theoretic one. For example, there is a sentence $\varphi$ in second-order logic which is valid iff the Continuum Hypothesis holds, and similarly for many other set-theoretic statements (this general phenomenon is reflected in the ludicrous size of the Hanf number of second-order logic).
Maybe this suggests that you should restrict attention to compact logics. If so, though, you'll be hard pressed to find examples other than first-order logic (and its sublogics) without invoking some set-theoretic ideas. This is because of Lindstrom's theorem, which states that there is no logic strictly stronger than first-order logic which is compact and has the Lowenheim-Skolem property. So, if you want a logic stronger than first-order logic, it had better do something complicated with respect to uncountable structures, specifically. An example is first-order logic with a quantifier for "there are uncountably many," which was shown to be (countably) compact by Keisler if I recall correctly; there are other more technical examples known.
