Tri-vector for Nambu bracket The Poisson bracket can be given via a bi-vector.
Is there a tri-vector for the Nambu 3-bracket? and higher?
 A: I assume we are talking about finite-dimensional smooth manifolds and using the Wikipedia definition of Nambu bracket.
A Nambu bracket is (in particular) a skew-symmetric multilinear multi-derivation $C^\infty(M) \times \cdots\times C^\infty(M) \to C^\infty(M)$.
Proof. Basically, if it is a derivation in the last argument, then skewness implies that it is a derivation in the other arguments as well.
To be precise, let $h_1,\ldots,h_n \in C^\infty(M)$ and suppose that $h_k = f_kg_k$ is a product. 
Choose $\sigma \in S_n$ such that $\sigma(n) = k$ and $\sigma(k) = n$ (e.g. a transposition or the identity). Then:
$$\begin{align*}\{h_1,\ldots,h_k,\ldots,h_n\} &= (-1)^{\epsilon(\sigma)}\{h_{\sigma(1)},\ldots,h_{\sigma(k)},\ldots,h_{\sigma(n)}\} \\ &= (-1)^{\epsilon(\sigma)}\{h_{\sigma(1)},\ldots,h_n,\ldots,f_kg_k\} \\ &= (-1)^{\epsilon(\sigma)}(\{h_{\sigma(1)},\ldots,h_n,\ldots,f_k\}g_k + f_k\{h_{\sigma(1)},\ldots,h_n,\ldots,g_k\}) \\ &= (-1)^{\epsilon(\sigma)}((-1)^{\epsilon(\sigma^{-1})}\{h_1,\ldots,f_k,\ldots,h_n\}g_k+ (-1)^{\epsilon(\sigma^{-1})}f_k\{h_1,\ldots,g_k,\ldots,h_n\}) \\ &= \{h_1,\ldots,f_k,\ldots,h_n\}g_k+f_k\{h_1,\ldots,g_k,\ldots,h_n\}\end{align*}$$
In the penultimate line we applied $\sigma^{-1}$ to the arguments of each Nambu bracket (at the price of a sign that will cancel against the one we had so far): the argument in the $k$th position is moved to the $n$th position and vice-versa.
In fact all skew-symmetric multilinear multi-derivations $C^\infty(M) \times \cdots\times C^\infty(M) \to C^\infty(M)$ arise from multi-vector fields on $M$ (there is a one-to-one correspondence). This is basically a consequence of Taylor's theorem; see e.g. [Poisson geometry
Lectures 1, 2, 3, Proposition 1.10].
In coordinates e.g. a $3$-bracket $\{f,g,h\}$ can be given via a 3-vector field $\sum_{i<j<k}\{x^i,x^j,x^k\}\frac{\partial}{\partial x^i}\wedge\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k}$.
