If t3suji wants to post an answer, then I will delete this answer. I am just posting this until further notice. There is a stratification of $X$ into locally closed subsets according to the dimension of the stabilizer group. To see this, apply Chevalley's result on upper semicontinuity of fiber dimension to the "inertia", i.e., the fiber product of the diagonal $\Delta:X\to X\times X$ and the morphism $\Psi:T\times X \to X\times X$, $\Psi(t,x) = (t\bullet x,x)$. So there is a dense open subset $X^o\subset X$ over which the stabilizer group is constant. Because subgroups of $T$ are discrete, there is a subgroup $T^o$ that acts trivially on $X^o$. Because $X^o$ is dense, $T^o$ acts trivially on all of $X$. Thus, without loss of generality, replace $T$ by $T/T^o$, and assume that $T$ acts freely on the dense open $X^o$.
Identify $T$ with the standard torus embedded in $\mathbb{P}^n$. The $T$-action on $\mathbb{P}^n$ has $n+1$ fixed points. For every $x\in X^o$, the $T$-action defines a $T$-equivariant rational transformation $f:\mathbb{P}^n \dashrightarrow X$ that is regular on $T$ and that sends $[1,\dots,1]$ to $x$. Denote by $\nu:Y\to \mathbb{P}^n$ the normalization of the closure of the graph of $f$ in $\mathbb{P}^n\times X$. Thus, there is an induced $T$-action on $Y$, and $\nu$ is $T$-equivariant, projective, and birational. Also denote by $g:Y\to X$ the induced morphism, so $f\circ \nu$ equals $g$ as $T$-equivariant rational transformations.
Every irreducible component of every fiber of $\nu$ over a fixed point is $T$-invariant and projective, thus it contains a fixed point. So the $T$-fixed points of $Y$ surject onto the $T$-fixed points of $\mathbb{P}^n$. The claim is that every pair of $T$-fixed points of $Y$ that are identified under $g$ are also identified under $\nu$. Assuming the claim, then $g(Y)$ contains at least $n+1$ $T$-fixed points.
The claim is proved by contradiction. The key is that, for every pair $y$ and $z$ of $T$-fixed points of $Y$ mapping to distinct $T$-fixed points of $\mathbb{P}^n$, there is a cocharacter $\lambda:\mathbb{G}_m\to T$ such that the orbit $\lambda(\mathbb{G}_m)\cdot [1,\dots,1]$ closes up to a $\mathbb{P}^1$ whose strict transform in $Y$ contains $y$ and $z$. So, if $y$ and $z$ are identified, then $g$ restricts on this rational curve to a $\lambda(\mathbb{G}_m)$-equivariant morphism that identifies precisely two points. Moreover, there is a $\lambda(\mathbb{G}_m)$-linearized ample invertible sheaf on this curve. But for a $\mathbb{P}^1$ with $0$ and $\infty$ identified, i.e., for a "nodal plane cubic", for every ample invertible sheaf, the group of automorphisms of the nodal curve together with the invertible sheaf is finite (essentially for the same reason as in the case of Abelian varieties). This contradicts that the automorphism group contains the infinite group $\lambda(\mathbb{G}_m)$. This contradiction proves that $y$ and $z$ are mapped to distinct points under $g$, even without the hypothesis that $X$ is normal.