Infinitary logics like L(𝜔1,𝜔) (the simplest and best-behaved of the bunch) and higher-order logics like second-order logic extend first-order logic in different directions. Infinitary logics extend first-order logic essentially by allowing "longer" formulas in controlled ways. On the other hand, roughly speaking the order of a logic refers to the sorts of things it can quantify over: e.g. can it quantify over merely elements of the structure, or subsets of the structure as well, or even more complicated configurations (e.g. sets-of-subsets, which would take us into third-order)?
Neither subsumes the other: for example, by a simple counting argument there is an L(𝜔1,𝜔)-sentence with no second-order equivalent (although there are some surprising subtleties), and one can also show with more work that there is a second-order sentence with no L(𝜔1,𝜔)-equivalent - for example, there is a second-order formula true in exactly those structures whose cardinality is a successor cardinal, and you can't do that with an infinitary sentence.
- Incidentally it turns out that infinitary logics are vastly better behaved in general than even second-order logic, but this gets a bit technical.
Now to your question proper, we definitely can have something like 𝜔th order L(𝜔1,𝜔), call it "𝜔L(𝜔1,𝜔)" for brevity. Basically, we just combine the syntactic operations of both the "L(𝜔1,𝜔)-part" and the "𝜔th-order" part! So for example we could have a disjunction of infinitely many clauses, with the nth clause being an nth-order first-order sentence.