If I understand your question correctly, the "easy" version of the question you ask is unknown in dimension $\ge 3$, and is basically easy for surfaces of interest.

These questions are certainly unknown in dimension $\ge 3$, or else the Sato-Tate conjecture for weight $>3$ would have been known $12$ months ago, rather than $4$ months ago.

I take it you are only asking that the action of Frobenius is neither almost always trivial ($T(p) = 0$) or acts as the identity ($T(p) = 1$), which is much weaker than asking that there exist any supersingular primes.
Assume otherwise. Fix a prime $\ell$, and Let $V$ be the $\ell$-adic etale cohomology
$H^2(X)$ with the usual action of $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Let $F_p(T)$ denote the characteristic polynomial of Frobenius. It has coefficients in $\mathbf{Z}[x]$. Say one wants to show that there exist infinitely many primes $p$ such that $T(p) \ne 0$. You are imposing the condition that $F_p(T) \equiv T^n \mod p$ for all but finitely many $p$, where $n = \mathrm{dim}(V)$.

Choose a prime $L > 2n$. There will a positive density of primes such that $F_p(T) \equiv (T-1)^n \mod L$.
If $p \equiv 1$ modulo $L$, These conditions together imply that the trace of $F_p(T)$, which is *a* *priori* an integer, is $np \mod pL$. By the Weil conjectures, the roots of $F_p(T)$ have absolute value $p$, and thus, since $L > 2n$, they must all equal $p$. By Cebatarev density, it follows that a finite index subgroup of the Galois group acts (on the semisimplification of $V$) via the cylotomic character on $V$. By looking at Hodge-Tate weights this implies that $h^{2,0} = h^{0,2} = 0$, and so (for example) one has a contradiction for a K3-surfaces.

If one wants to rule out that $T(p) = 1$ for (almost all) $p$, one is imposing the condition that $F_p(T) = (T-1)^n \mod p$. In the same way one obtains an open subgroup of the Galois group such that the trace of Frobenius is always $n$, and then one deduces that $h^{0,2} > 0$ and $h^{2,0} = 0$, which can never happen.

EDIT: Forgot to mention that it was Ogus who proved in ~70's that Abelian surfaces had infinitely primes of ordinary reduction, presumably by a very similar argument. Well, except for the comparison theorem of Faltings I used above...