Editet question
Let $(M, \pi)$ be an oriented Poisson manifold. Is there a natural way to build a volume form from the Poisson bivector $\pi$?
Original question
It is always possible to build a volume form on a symplectic manifold; for example if $(M, \omega)$ is a $2n$-dimensional symplectic manifold, then $\Omega = \underbrace{\omega \wedge \dots \wedge \omega}_{n\text{ times}}$ is a volume form. Is there an analogue way to orient Poisson manifolds?
Let $(\mathcal M,\pi)$ be an orientable Poisson manifold.
A natural volume form could be probably intended as: every Hamiltonian vector field preserves this volume form.
In this acception, Poisson-Lichnerowitz (PL) cohomology (often called just Poisson cohomology) provides a criterion. I am not an expert in the subject, so I am essentially digesting what I read on Wikipedia.
Let us recall the definition of the first PL cohomology group, denoted $H^1(\mathcal M,\pi)$. It is the quotient of the group of Poisson vector fields (by definition, those $X$ that preserve the Poisson structure, i.e., ${\rm Lie_X}\pi =0$) over the subgroup of Hamiltonian vector fields (by definition, those $X=X_H$ for some $H\in\mathscr C^{\infty}(M)$, where $X_H(f)= \lbrace f,H\rbrace$ for all $f\in\mathscr C^{\infty}(M)$). (Recall that every Hamiltonian vector field is a Poisson vector field.)
Fix a volume form $\lambda$ and recall that any volume form on a manifold allows for the definition of divergence (a scalar denoted ${\rm div}_\lambda\,X$) of a vector field $X$, namely by requiring the identity $$ {\rm Lie_X}\lambda = ({\rm div}_\lambda\,X) \lambda. $$ (A nice exercise is to check that this reproduces the standard notion when the manifold is an open set in $\mathbb R^n$ and $\lambda=dx^1\wedge\dots\wedge dx^n$.)
Define, for a fixed $\lambda$ volume form, the Poisson modular vector field $M_\lambda$ by its action as a derivation on functions: $$ M_\lambda:f\mapsto {\rm div}_\lambda\,X_f. $$
Exercise 1. $M_\lambda$ is a Poisson vector field.
Exercise 2. The difference $M_{\lambda_1}-M_{\lambda_2}$ is an Hamiltonian vector field for any pair of volume forms $\lambda_1,\lambda_2$.
(This allows to regard $[M_\lambda]\in H^1(\mathcal M,\pi)$.)
Exercise 3. $[M_\lambda]=0$ if and only if there exists a volume form $\lambda$ such that $\lambda$ is preserved by all Hamiltonian vector fields.
More details can be found in the above Wikipedia page (and probably, in the references therein).