Consider a Hamiltonial $\mathbb T ^m$-action on a compact connected symplectic manifold $M$. Let $\mu : M \to \mathbb R ^m$ be the momentum map, $\mu= (\mu_1, \dots, \mu_m)$.
Assuming that the action is effective (ie, all elements except the identity move at least one point in $M$), I want to show that there is a point $p \in M$ such that $$\text{d}_p \mu_1, \dots, \text{d}_p \mu_m$$ are linearly independent.
Remark. I wanted to proof a Corollary of the well-known Atiyah-Guillemin-Sternberg theorem, that states that in the conditions of the questions (ie, of the theorem), the torus action has $m+1$ fixed points. But the linearly independence condition is the one that I miss.
I prefer to work in a coordinate-free fashion, as I consider this ultimately less confusing, so let $\mathfrak{t}$ denote the Lie algebra of $G:= \mathbb{T}^m$ and $\mathfrak{t}^*$ its dual.
Let $(M, \omega)$ be a compact symplectic manifold and let $\mu : M \to \mathfrak{t}^*$ be 'the' moment map associated to a Hamiltonian $G$-action on $(M, \omega)$. You are seeking for the existence of a point $p \in M$ such that the differential $d\mu_p : T_pM \to \mathbb{R}^m$ is surjective.
The effectiveness of this Hamiltonian toric action not only implies the existence of such a point $p \in M$, but also that the set of such points is an open dense subset of $M$.
That this set is open follows readily from the fact that the condition "$d\mu_p : T_pM \to \mathbb{R}^m$ is surjective" on $p$ is open. This part does not require faithfulness of the action.
Let us now rephrase the surjectivity condition you are interested in. For $p \in M$, let $X_p : \mathfrak{t} \to T_pM : \xi \mapsto X_p(\xi) := (X_{\xi})(p)$ where $X_{\xi}$ is the Hamiltonian vector field associated to the Hamiltonian function $H_{\xi}(p) := \langle \mu(p), \xi \rangle$. From $X_p(\xi) \lrcorner \, \omega_p = - (dH_{\xi})_p = - \langle d\mu_p, \xi \rangle$ and the nondegeneracy of $\omega$, it follows that $\mathrm{ker} \, X_p \subset \mathfrak{t}$ is the annihilator of the set $\mathrm{im} \, d\mu_p \subset \mathfrak{t}^*$. Consequently, $d\mu_p$ is surjective is and only if $X_p$ is injective, that is if and only if the stabilizer subgroup $G_p \le G$ is discrete (which is to say that its Lie algebra is just $\{0\} \subset \mathfrak{t}$).
Therefore you are interested in the set of points $p \in M$ whose stabilizer is zero-dimensional, that is the complement of those points $q \in M$ whose stabilizer is at least one-dimensional. As every connected component of a Lie group has the same dimension, we can only consider below connected subgroups of dimension at least one. Notice also that a stabilizer subgroup is closed.
Let $K < G$ be a connected closed subgroup of dimension at least 1. It is thus comppact, but it is also Lie. From these properties, there is a nonzero $\xi \in Lie(K)$ generating a one-parameter subgroup of $K$ which is dense in $K$. (Indeed, as $K$ is a compact abelian Lie group, it is a torus, so it is a product of a number of circles of some periods and it suffices to choose $\xi$ with components which are incommensurable with those periods.) But $\xi$ acts on $M$ through the Hamiltonian $H_{\xi}$. It follows that the points $q \in M$ fixed by $K$ are the same as the critical points of $H_{\xi}$, a crucial observation.
Another crucial observation of Atiyah and of Guillemin-Sternberg is that for a Hamiltonian torus action, each Hamiltonian $H_{\xi}$ is either constant or a Morse-Bott function on $M$. In the latter case, the set of regular points of $H_{\xi}$ is an open dense subset of $M$.
If one assumes that the action is effective, then $H_{\xi}$ is not constant for $\xi \neq 0$. Combining this with the two previous paragraphs, we conclude that for any closed subgroup $K < G$ of dimension at least 1, the set of points in $M$ which do not have $K$ in their stabilizer is an open dense set of $M$.
To conclude the argument, we need one last ingredient, which is to show that the collection of subgroups $K < G$ of dimension at least 1 appearing as stabilizers of the action is countable. Indeed, it would follow that the set of points $p \in M$ whose stabilizer is discrete is the intersection of countably many open dense sets, and would thus be itself dense according to the Baire category theorem.
To show that this collection is countable, there is two instructive ways:
1) Recall that stabilizers are compact Lie subgroups and have as such finitely many connected components. The number of closed connected subgroups $K < G$ is countable; this follows from arguing on the possible closure of one-parameter subgroups of $G$. Thus stabilizers subgroups belong to a countable family.
2) A result related to what Guillemin-Sternberg call the Koszul-Mostow theorem in their book Symplectic techniques in physics states that for a smooth group action by a compact Lie group on a compact manifold, the number of conjugacy classes of stabilizer subgroups is finite. When the group is abelian, it follows that the number of stabilizers is finite.