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preliminary remark

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

the statement

I claim that one can prove the following 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 $x$.

Tell me if I have misunderstood the question.

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

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

2. 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.

preliminary remark

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

the statement

I clam that one can prove the following 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 $x$.

Tell me if I have misunderstood the question.

preliminary remark

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

the statement

I claim that one can prove the following 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 $x$.

Tell me if I have misunderstood the question.

Isn't it a consequence of the existence of action-angle semi-global coordinates? Namely, one has that

If $F=(f_1,\dots,f_n)$ is proper, then for any lagrangian invariant torus $\Lambda$ there exists a neighborhood $U$ of $\Lambda$ that is symplectomorphic to
a neighborhood $V$ of a fiber of the projection in the cotangent bundle $T^*\mathbb{T}^n$ of the $n$-dimensional torus $\mathbb{T}^n$ (namely, $V=\{(\theta,I), |I-I_0|\leq r\}$).

It seems to me that the zero section $\{\theta=0\}$ satisfies your requirement.

EDIT: if you read french then you can find a proof of this statement in Lecture Notes by Colin de Verdière (proof of Theorem 20, pages 50-51).

EDIT 2: I think the proof is the same as the one in Colin de Verdiere's notes. Namely, you first prove it in the dimension $1$ situation. Here you don't need properness assumption since the conclusion you are asking for is only a local statement and not a semi-global one (it is about the neighborhood of a point, while action-angle is about the neighborhood of a Lagrangian leaf). The rest of the proof is the same except that instead of a free action of $\mathbb{T}^n$ you have a locally free action of $\mathbb{R}^n$ (which is the only thing you need for your conclusion).

Obvious remark concerning vocabulary: there is one more thing to care about. It is what people mean by $f_1,\dots,f_n$ being independant. I personnaly mean that at almost any point their derivatives are linearly indpendant. Therefore in this case the conclusion is true for almost all points in $M$. At critical points of $F$ this is very easy to construct a counter-example in dimension $2$: take $F$ to be the square of the norm in $\mathbb{R}^2$, and look at what happens at the origin.

preliminary remark

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

the statement

I clam that one can prove the following 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 $x$.

Tell me if I have misunderstood the question.