### preliminary remark

I assume that being independant for functions here means that their differentials at **any** point are linearly indenpendant (and not at **almost** any point, like it is sometimes assumed... in which case a counter-example is easy to find for $n=2$: take $f_1$ to be the square of the norm on $\mathbb{R}^2$, and look at the origin).

### the statement

I claim that one can prove the following weak version of the action-angle coordinate Theorem, under the hypothesis of the question:

For any $x\in M$ there exists Darboux coordinates $(p_1,\dots,p_n,q_1,\dots, q_n)$

around $x$ such that the leaves of the Lagrangian foliation are $(q_1,\dots,q_n)=cst$.

Then the local manifold $\{p_1=\dots=p_n=0\}$ satisfies your requirement.

To prove the statement, consider $q_i=f_i-f_i(x)$ and extend them to a Darboux chart $(q_1,\dots,q_n,p_1,\dots,p_n)$ around $x$.

### source of confusion

foliations may be very wilde... but here we actually have a submersion.

in usual action-angle coordinates Theorem on needs some properness assumption. But the usual Theorem tell us about properties of semi-global coordinates. Here we were dealing with a purely local statement and we don't need any properness assumption.