The Poisson geometry is, I think, a red herring. I will explain in this answer how to construct $\Omega^{2n}(X)$ as a $C^{\infty}(X)$ module, and how to construct the volume form $\omega^{n}$ within it. So I think the right question to ask is:

Let $X^m$ be a smooth, orientable, compact manifold. Suppose that we are given $\Omega^{\bullet}(X)$ with its dga structure. How can we recover the functional $\int_X: \Omega^m \to \mathbb{R}$?

Given $C^{\infty}(X)$, we can construct $T^1(X)$ as the space of $\mathbb{R}$-linear derivations $C^{\infty}(X) \to C^{\infty}(X)$. Then we can construct $\Omega^1(X)$ as $\mathrm{Hom}_{C^{\infty}(X)}(T^1(X), C^{\infty}(X))$. (Attempting to construct $\Omega^{1}(X)$ more directly runs into problems.) We then have the map $d: C^{\infty}(X) \to \Omega^1(X)$ which, given a function $f \in C^{\infty}$, produces the map $Y \mapsto Y(f)$ in $\mathrm{Hom}_{C^{\infty}(X)}(T^1(X), C^{\infty}(X))$. The Serre-Swan correspondence then lets us build the sections of any vector bundle on $X$ which is obtained from tensor powers of $\Omega^1(X)$ by applying the corresponding tensor operations on $C^{\infty}(X)$-modules. For example, $(\bigwedge^2 T^1)(X)$ is the antisymmetric elements of $T^1(X) \otimes_{C^{\infty}(X)} T^1(X)$.

In particular, we can talk about $\bigwedge^2 T^1(X)$, and we have the evaluation pairing $\bigwedge^2 T^1(X) \times C^{\infty}(X) \times C^{\infty}(X) \to C^{\infty}(X)$. So we can talk about the unique bivector $\eta$ in $\bigwedge^2 T^1(X)$ which realizes the Poisson bracket. Also, we can take about the map $(\bigwedge^2 T^1(X))^{\otimes n} \to \bigwedge^{2n}(T^1(X))$. So we can produce the form $\eta^{\otimes n}$.

We can say that the manifold $X$ is symplectic if and only if $\eta^{\otimes n}$ is a generator of the $C^{\infty}(X)$-module $\bigwedge^{2n} T^1(X)$. If so, then $\omega^n$ is the reciprocal element in $\Omega^{2n}(X)$.

All of the above, though a pain, is definitely algebraic. So I claim that we are reduced to the boxed question.

I have a pretty elegant answer to the boxed question for circles, and a not very elegant answer for general $m$. I hope this question will attract some better answers.

Let $X$ be a circle, and let $\eta \in \Omega^1(X)$. We will say that $\eta$ is a resonant class if there are $x$ and $y \in C^{\infty}(X)$ such that
$$x^2+y^2=1,\ dx = y \eta,\ dy = - x \eta.$$
The point is that, if these equations are solvable, then we can locally write $(x,y)$ as $(\cos \theta, \sin \theta)$ where $d \theta = \eta$.

So these equations are globally solvable if we can find a global such $\theta$, modulo $2 \pi$. In other words, they are globally solvable if and only if $\int_X \eta \in 2 \pi \mathbb{Z}$. So this gives us a description of the map $\int_X$ up to sign; it takes the resonant classes in $\Omega^1(X)$ to $2 \pi \mathbb{Z}$.

Here is my bad answer to the general case. We will say that $\eta$ "comes from a sphere" if there are functions $x_0$, $x_1$, ..., $x_m$ on $X$ such that $\sum x_i^2=1$ and $\eta = \sum (-1)^r x_r dx_0 \wedge dx_1 \wedge \cdots \widehat{dx_r} \cdots \wedge dx_m$. In other words, $\eta$ is pulled back from the volume form on $S^m$ along a smooth map.

Then we can describe $\int_X$ by the two normalizations that (1) $\int_X$ is $0$ on $d \Omega^{m-1}(X)$ and (2) $\int_X$ takes integer multiples of the area of $S^m$ on those forms that come from a sphere. The normalization is correct because any compact oriented connected $m$-fold has a smooth degree $1$ map to an $m$-sphere. (Proof on request.)

I say this is ugly for two reasons. First, the construction of that degree $1$ map is a partition of unity argument. So I suspect that this description will be completely useless in practice.

Second, although every integral cohomology class is pulled back from a sphere, I don't think that every integral volume form is. So I really need both conditions in the normalization above. That strikes me as inelegant.