I don't know how Tate did it but here is one way. Let $\zeta$ be such that $\zeta^{q+1}=-1$ and put $a_j=(0\colon\cdots\colon1\colon\zeta:\cdots\colon0)$, $j=0,\ldots,i$ with the $1$ in coordinate $2j$. Then the linear span $L$ of these points is contained in the Fermat hypersurfaces and gives a subvariety of middle dimension. I claim that its class is not a multiple of the $i+1$'st power of the hyperplane section. Indeed, transform $L$ by the automorphism permuting the two first coordinates to get $L'$. Then (assuming that $q$ is odd say) $L$ and $L'$ are disjoint so that $[L]\cdot [L']=0$ but if the class of $L$ would be a multiple of the linear subspace section this cannot be. Hence, $[L]$ projects non-trivially to the non-trivial irreducible representation.
Note that Shioda et al have studied cycles on general Fermat hypersurfaces and verified the Tate conjecture in many (all?) cases.
Addendum: It is tricky to get dimensions and stuff correct so let me expand upon this in the simple case when $i=1$ and $q=2$ (this is a classical example appearing for instance on pp. 176-177 of Mumford: Algebraic Geometry I Complex projective varieties -- everything works in characteristic different from $3$). Then we get the line $(a\colon a\zeta\colon b\colon b\zeta)$ and by letting the monomal automorphisms of the surface act we get all $27$ lines on the Fermat cubic. It is of course true that the line is the intersection of the cubic and a linear subspace but the intersection is not proper so the class of $L$ is not a power of the hyper plane section.
Note incidentally that in comments (p. 180) Mumford points the special nature of characteristic $2$ for this example essentially saying that there an index $2$ subgroup of the automorphism group of the Néron-Severi lattice preserving the canonical class can be realised by automorphisms of the surface.
