I see where is the problem.

First of all, the right statement of Tate conjecture is for $X$ over $\mathbb{F}_q$ (not over $\overline{\mathbb{F}_q}$!), and Tate conjecture predicts:

$c_1: Pic(X)\otimes\mathbb{Q}_p\rightarrow H^2_{crys}(X/W(\mathbb{F}_q))^{\Phi=p}\otimes\mathbb{Q}_p$ is an isomorphism.

Secondly, Dieudonn$\acute{e}$-Manin classification only works over algebraically closed fields like $\overline{\mathbb{F}_q}$, doesn't work over $\mathbb{F}_q$, and when we talk about slopes of an F-crystal $H$ over $\mathbb{F}_q$, we actually mean the slopes of $H\otimes W(\overline{\mathbb{F}_q})$.

Now assume the K3 crystal $H^2_{crys}(X/W(\mathbb{F}_q))$ has slopes $1-\frac{1}{h},1,1+\frac{1}{h}$, and assume it's non-supersingular.

The fact is, although the rank of $(H^2_{crys}(X/W(\mathbb{F}_q))\otimes W(\overline{\mathbb{F}_q}))^{\Phi=p}$ is indeed $22-2h$, the rank of $H^2_{crys}(X/W(\mathbb{F}_q))^{\Phi=p}$ can be smaller:

Here is an easy example of F-crystal over $\mathbb{F}_9$ of slope 1 rank 1 but $\Phi=p$ space is trivial:

Since $W(\mathbb{F}_9)\cong \mathbb{Z}_9$ there is an 8-th roots-of-unity $\zeta_8$ in $W(\mathbb{F}_9)$, let $H$ be the rank 1 $W(\mathbb{F}_9)$-module where $\Phi$ maps $1$ to $\zeta_8p$.

Then this F-crystal maps any $x\in W(\mathbb{F}_9)$ to $\zeta_8p\sigma(x)$, where $\sigma\in Gal(\mathbb{Q}_9/\mathbb{Q}_3)$ is the only non-trivial (Frobenius) map.

Then no element in $\mathbb{Z}_9$ satisfy $\zeta_8p\sigma(x)=px$. (This is simply because the multiplicative subgroup of $\mathbb{Q}_9^{\times}$ generated by elements of the form $\sigma(x)/x$ is actually $\mu_4$, the group of 4th-roots of unities-this is an easy computation.)

But after base change to $W(\overline{\mathbb{F}_9})$, suppose $\zeta_{16}$ is a square root of $\zeta_8$, let $x=\zeta_{16}^7$, then $\Phi(x)=px$ which means $H$ indeed has slope $1$.

I see where is the problem.

First of all, the right statement of Tate conjecture is for $X$ over $\mathbb{F}_q$ (not over $\overline{\mathbb{F}_q}$!), and Tate conjecture predicts:

$c_1: Pic(X)\otimes\mathbb{Q}_p\rightarrow H^2_{crys}(X/W(\mathbb{F}_q))^{\Phi=p}\otimes\mathbb{Q}_p$ is an isomorphism.

Secondly, Dieudonn$\acute{e}$-Manin classification only works over algebraically closed fields like $\overline{\mathbb{F}_q}$, doesn't work over $\mathbb{F}_q$, and when we talk about slopes of an F-crystal $H$ over $\mathbb{F}_q$, we actually mean the slopes of $H\otimes W(\overline{\mathbb{F}_q})$.

Now assume the K3 crystal $H^2_{crys}(X/W(\mathbb{F}_q))$ has slopes $1-\frac{1}{h},1,1+\frac{1}{h}$, and assume it's non-supersingular.

The fact is, although the rank of $(H^2_{crys}(X/W(\mathbb{F}_q))\otimes W(\overline{\mathbb{F}_q}))^{\Phi=p}$ is indeed $22-2h$, the rank of $H^2_{crys}(X/W(\mathbb{F}_q))^{\Phi=p}$ can be smaller:

Here is an easy example of F-crystal over $\mathbb{F}_9$ of slope 1 rank 1 but $\Phi=p$ space is trivial:

Since $W(\mathbb{F}_9)\cong \mathbb{Z}_9$ there is an 8-th roots-of-unity $\zeta_8$ in $W(\mathbb{F}_9)$, let $H$ be the rank 1 $W(\mathbb{F}_9)$-module where $\Phi$ maps $1$ to $\zeta_8p$.

Then this F-crystal maps any $x\in W(\mathbb{F}_9)$ to $\zeta_8p\sigma(x)$, where $\sigma\in Gal(\mathbb{Q}_9/\mathbb{Q}_3)$ is the only non-trivial (Frobenius) map.

Then no element in $\mathbb{Z}_9$ satisfy $\zeta_8p\sigma(x)=px$. (This is simply because the multiplicative subgroup of $\mathbb{Q}_9^{\times}$ generated by elements of the form $\sigma(x)/x$ is actually $\mu_4\times(\frac{4+3i}{5})^{\mathbb{Z}_3}$, the group of 4th-roots of unities-this is an easy computation.)

But after base change to $W(\overline{\mathbb{F}_9})$, suppose $\zeta_{16}$ is a square root of $\zeta_8$, let $x=\zeta_{16}^7$, then $\Phi(x)=px$ which means $H$ indeed has slope $1$.