Let $k$ be a field of characteristic $0$. If $m\ge2$, I denote $P_m$ the *standard polynomial* in $m$ non-commutating indeterminates:
$$P_m(X_1,\dotsc,X_m)=\sum_{\sigma\in\mathfrak S_m}\epsilon(\sigma)X_{\sigma(1)}\dotsm X_{\sigma(m)}.$$
We say that a $k$-algebra $A$ is $m$-commutative if $P_m$ is an identity over $A$, that is
$$\forall a_1,\dotsc,a_m\in A,\qquad P_m(a_1,\dotsc,a_m)=0,$$
and $P_{m-1}$ is not. Remark that $2$-commutativity is just commutativity. 

Amitsur–Levitzki's Theorem is that $M_n(k)$ is $(2n)$-commutative.

> My question is whether there exist (interesting) algebras that are $m$-commutative for an odd $m$. For instance, what would be an example of a $3$-commutative algebra ?

**Edit**. As [mentioned](https://mathoverflow.net/questions/430620/is-there-a-3-commutative-algebra#comment1108101_430620) by *user49822*, $A$ may not be unital, otherwise $(2k+1)$-commutativity implies $(2k)$-commutativity by specifying $a_m=1$. Thus there remains the question of whether a non-unital algebra can be $3$-commutative.