This special case of the problem of estimating the number of ${\bf F}_p$-points on a variety is much easier than the Weil bound: the standard estimates on Gauss sums suffice. I hope that this is elementary enough. I recite at the end of this answer a derivation of a bound that generalizes to arbitrary "diagonal hypersurfaces" and reduces the present problem to an easy finite computation (through $p=73$).

With more precise information about quartic Gauss sums (somewhat less elementary, but still known to Gauss — it's in an appendix to *Disquisitiones Arithmeticæ*), one can obtain an exact formula for the number of rational points. Assume that $p \equiv 1 \bmod 4$, else the count is $(p+1)^2$ and is entirely elementary (we may as well count on the quadric $w^2+x^2+y^2+z^2=0$ because in this case every number mod $p$ has as many square roots as it has fourth roots). Write $p = a^2 + b^2$ with $a$ odd. Then the number of rational points is $p^2+mp+1+4a^2$, where $m=18$ or $-6$ according as $p$ is congruent to $1$ or $5 \bmod 8$. This shows that $p=5$ is the only case where there are no rational points. The formula also fits in with the fact that the diagonal Fermat quartic is a K3 surface of maximal Picard number (= Néron-Severi rank), so I'll add the "k3-surfaces" tag.

To connect the enumeration problem with Gauss sums we argue as follows. Again it is enough to consider the case $p \equiv 1 \bmod 4$, since if $p \equiv 3 \bmod 4$ the problem is equivalent to the easier task of enumerating rational points on $w^2+x^2+y^2+z^2=0$, for which the same technique works (using only quadratic Gauss sums) but is overkill.

For $n \in \bf Z$ write $e(n) = \exp(2\pi i n /p)$. Then the number of solutions of $w^4+x^4+y^4+z^4 = 0$ in ${\bf F}_p$ is $p^{-1} \sum_{k=0}^{p-1} S_k^4$ where $S_k = \sum_{t=0}^{p-1} e(kt^4)$. [Proof: expand $\sum_{k=0}^{p-1} S_k^4$ as
$$
\sum_{k=0}^{p-1} \left[ \mathop{\sum\sum\sum\sum}_{w,x,y,z\phantom0\bmod \phantom0p}
\phantom. e(k(w^4+x^4+y^4+z^4))\right]
$$
and switch the order of summation: the sum over $k$ is $p$ if $w^4+x^4+y^4+z^4 = 0$,
and zero otherwise. NB we're counting affine solutions, including $(w,x,y,z)=(0,0,0,0)$, not projective solutions.]
Now $S_0 = p$. I claim that $|S_k| \leq 3\sqrt{p}$ for $k \neq 0$. It will follow that the number of solutions is within $81(p-1)p^2$ of $p^4$. In particular the number of solutions exceeds $1$ once $p \geq 81$, so the number of projective ${\bf F}_p$-points is positive. Checking the remaining cases $p=13,17,29,37,41,53,61,73$ is routine.

To prove that $|S_k| \leq 3 \sqrt{p}$ for $k \neq 0$, write $S_k = \sum_{n=0}^{p-1} c_n e(kn)$ where $c_n$ is the number of solutions of $t^4 \equiv n \bmod p$. Now $c_n = 1 + \chi(n) + \chi^2(n) + \chi^3(n)$ where $\chi$ is a Dirichlet character mod $p$ of order $4$. Thus
$$
S_k(n) = \sum_{j=0}^3 \left[ \sum_{n=0}^{p-1} \phantom. \chi^j(n) \phantom. e(kn) \right].
$$
For $j=0$, the inner sum vanishes; for $j = 1,2,3$ it's a Gauss sum, which has absolute value $\sqrt p$. Hence ${}$$|S_k| \leq \sum_{j=1}^3 \sqrt p = 3 \sqrt p$ and we're done.

I'll leave the proof of the formula $p^2+mp+1+4a^2$ as an "exercise" because this answer is already rather long...