This answer may be a bit anticlimactic, but you can take free associative algebra without $1$ (same as ideal of positive degree elements in tensor algebra) and look at its quotient by *all* products of degree 3 (or any other $n$). It's obvious that it will be strictly $n$-commutative in your sense. Study of properties and representations of such *nil-rings* was quite popular few decades ago, but now mostly fell into obscurity.

Similarly you can take an ideal generated in free nonunital algebra by all substitutions of its elements into $n$-commutativity relation; it will be a free algebra in a variety defined by that relation. That varieties do not coincide, as witnessed by example above, so their free algebras would not be isomorphic as well.