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.