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I have just thought of another interesting example, which is a bit peculiar, in that the 3-commutativity property arises for a natural subspace of an algebra which itself does not satisfy any identity. Specifically, consider the algebra $D_1$ of polynomial differential operators on the line; it consists of linear combinations of elements $x^i\partial^j$, where $\partial x - x\partial =1$ holds. It is well known that this algebra does not satisfy any polynomial identity. However, let us consider the subspace (in fact, a Lie subalgebra) $W_1$ of this algebra consisting of all vector fields; it consists of linear combinations of elements $x^i\partial$. I claim that any three elements of this subspace 3-commute (in some papers, this is described by saying that 3-commutativity is a weak identity of $D_1$ with respect to its Lie subalgebra $W_1$). This follows from the fact that $$ f\partial g\partial h\partial=f\partial g (h\partial + h')\partial= f(gh\partial + g'h+gh')\partial^2+f(gh'\partial+gh''+g'h')\partial, $$ which can be written as $$ fgh\partial^3+(fg'h+2fgh')\partial^2+(fgh''+fg'h')\partial, $$ and this is clearly killed by total anti-symmetrization in $f,g,h$, and does not follow from any stronger identity.