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.

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 easy to see that obtained algebra will be strictly $n$-commutative in your sense (it's almost obvious: tensor algebra is strictly graded). 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; quotient of tensor algebra by constructed ideal 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.