This is true in a much more general setting. Let $X$ be any normal projective $T$-variety defines over an algebraically closed field. Then I claim that $\# X^T>\dim X$. We show this in two steps.
(1) According to a theorem of Sumihiro, there is an equivariant embedding of $X$ into a projective space ${\bf P}^n$. Normality is used only in this first step. Often this step is superfluous because $X$ is sitting already in a projective space. Then all we need for the second step is that $X$ is closed.
(2) Now let $T$ act linearly on ${\bf P}^n$ and let $X\subseteq{\bf P}^n$ be closed and $T$-stable. Then there are three substeps:
(a) Assume $X={\bf P}^n$. Then ${\bf P}^n={\bf P}(V)$ where $V$ is an $n+1$-dimensional representation of $T$. This representation is diagonalizable. Hence $V=U_1\oplus\ldots\oplus U_n$ where each $U_i$ is $1$-dimensional with $T$ acting with a character $\chi_i$. Then $T$ has $n+1$ fixed points namely ${\bf P}(U_0),\ldots,{\bf P}(U_n)$.
(b) Assume $X^T=({\bf P}^n)^T$. Then $\# X^T=\#({\bf P}^n)^T>n\ge\dim X$ by (a).
(c) Finally assume $X^T\subsetneq ({\bf P}^n)^T$. Then there is a $T$-fixed point $x$ not lying in $X$. Let $\pi:{\bf P}^n\setminus\{x\}\to{\bf P}^{n-1}$ be the projection with center $x$. Then the restriction of $\pi$ to $X$ is both affine and projective hence finite. Let $\tilde X:=\pi(X)\subseteq{\bf P}^{n-1}$. Then $\#\tilde X^T>\dim\tilde X=\dim X$ by induction on $n$. We conclude with $\# X^T\ge\#\tilde X^T$ since the fibers of $\pi|_X$ are finite and $T$ is connected.
Edit: Part (2) is Prop. IV.13.5 of A. Borel, Linear Algebraic Groups 2nd ed., Springer GTM 126.