Is there a proof of Warning's Second Theorem using p-adic cohomology? Let $\mathbb{F}_q$ be a finite field, $n \in \mathbb{Z}^+$, and $f_1,\ldots,f_r \in \mathbb{F}_q[t_1,\ldots,t_n]$ with $\operatorname{deg}(f_i) = d_i$.  Put $d = \sum_{i=1}^n d_i$ and suppose $d< n$. Let $V(f_1,\ldots,f_r) = \{ x \in \mathbb{F}_q^n \mid f_1(x) = \ldots = f_r(x) = 0\}$.  
The papers

C. Chevalley, Demonstration d’une hypothese de M. Artin Abh. Math. Sem. Univ. Hamburg 11 (1935), 73–-75

and 

E. Warning, Bemerkung zur vorstehenden Arbeit von Herrn Chevalley.  Abh. Math. Sem. Hamburg 11 (1935), 76–-83

contain the following results:
Chevalley's Theorem: We have $\# V(f_1,\ldots,f_r) \neq 1$.  
Warning's First Theorem: We have $\operatorname{char}(F) \mid \# V(f_1,\ldots,f_r)$.  
Warning's Second Theorem: If $V(f_1,\ldots,f_r)$ is nonempty, its cardinality is at least $q^{n-d}$.
The first two results are often combined as the "Chevalley-Warning Theorem", and the last result seems to be too often overlooked.  The Chevalley-Warning Theorem is the beginning of a long story of p-adic estimates on the number of $\mathbb{F}_q$-rational points on such $\mathbb{F}_q$-schemes, culminating in the Ax-Katz Theorem which determines the minimal $p$-adic valuation of $\# V(f_1,\ldots,f_r)$ as $f_1,\ldots,f_r$ range over all polynomials of degrees $d_1,\ldots,d_r$.  Famously, Katz used p-adic cohomology to prove his theorem.  Other, more elementary proofs have since been given, but this approach remains a very natural one.

Question: Is there a proof of Warning's Second Theorem using $p$-adic cohomology?  Does such a proof appear in the literature?  

Warning's proof of his second theorem is an ingenious elementary argument of the sort that nowadays goes under the moniker "Polynomial Method".  See this paper of Heath-Brown which includes Warning's proof as a point of departure to further work.  (To the best of my knowledge, Heath-Brown's paper is the only one which directly follows up on Warning's Second Theorem.  I would be very interested to learn of others.)
A more modern Polynomial Method approach to Warning's Second Theorem which leads to a different kind of generalization is discussed here.
Added: To address Daniel Litt's question raised in the comments: I do not know how to deduce Warning's Second Theorem (or even his first, for that matter) from the Weil Conjectures.  The fact that we are making no "geometric" assumptions on our affine subscheme would make such a deduction a forbidding endeavor for me....but I do not claim that it cannot be done.  I would be happy to broaden the question to ask about a cohomological proof of Warning II: you get to pick the cohomology theory.
 A: I do not think that there is such a proof in the litterature, but as you have guessed, $p$-adic cohomology has something to do with it.
Let $X=V(f_1,\dots,f_r)$. Let $K$ be the field of fractions of the Witt vectors of $k$, and let $\phi$ be the unique automorphism of $K$ such that $\phi(a)\equiv a^p \pmod p$. The rigid cohomology groups with compact supports of $X$, $H^i_{\rm rig,c}(X/K)$, are 
$\phi$-isocrystals, namely finite dimensional $K$-vector spaces  endowed with a $\phi$-linear automorphism $\Phi$.
If $k=\mathbf F_q$, with $q=p^r$, the Lefshetz trace formula in rigid cohomology asserts that 
$$ \# X(k)  = \sum (-1)^i \mathop{\rm Tr}(\Phi^r|H^i_{\rm rig,c}(X/K)). $$
(Note that $\Phi^r$ is linear, so that the trace makes sense.)
The theory of slopes of $\Phi$-isocrystals shows that Warning's theorem would follow from the condition that all slopes of $H^i_{\rm rig,c}(X/K)$ 
are $\geq 1$. As you said, it does not seem to be known without any smoothness assumption. Moreover, there might be cancellations in the trace formula that make this expectation wrong (see however this paper of Esnault and Katz).
However, there is a nice theorem of Berthelot, Esnault and Rülling in this direction: Let $X$ be a flat, regular and proper $R$-scheme whose generic fiber $X_K$ is geometrically connected and satisfies $H^i(X_K,\mathscr O_{X_K})=0$ for $i>0$. Then $\# X(k)\equiv 1\pmod q$.  The hard part of the proof consists in proving that the Witt vector cohomology vanishes in positive degrees.
A: This isn't quite what you're asking for, but I think it's some kind of prototype, so I hope you'll indulge me. 
Let me give a proof of the Chevalley-Warning Theorem (in the case the $f_i$ are homogeneous) via coherent cohomology of $\mathcal{O}_X$ (which if you'd like can be viewed as part of algebraic de Rham cohomology, which is the prototype for $p$-adic cohomology theories such as crystalline cohomology).
Namely, Fulton's trace formula (proven here, with I think a slight gap in the computation of $K^F(\mathbb{P}^n)$, and with the gap filled here or here) tells us that if $X$ is a projective variety over $\mathbb{F}_q$, with $q=p^k$, $$\# X(\mathbb{F}_q)=\sum_{i=0}^{\text{dim}(X)} (-1)^i Tr(\text{Frob }\mid H^i(X, \mathcal{O}_X))\bmod p.$$
Now suppose your polynomials are homogeneous.  The degree conditions on your polynomials should imply that the projective variety $X$ they define has $H^i(\mathcal{O}_X)=0$ for $i>0$, so the right hand side is just $1$.  So $$\# X(\mathbb{F}_q)=1\bmod p.$$
Thus the affine variety they define has $(q-1)(\#X(\mathbb{F}_q))+1$ points, which is divisible by $p$. $\blacksquare$
By the way, for smooth varieties, one may prove Fulton's trace formula via algebraic de Rham cohomology.  Namely, one copies the usual proof of Lefschetz fixed point for algebraic de Rham cohomology, which gives a similar trace formula for $\#X(\mathbb{F}_q)\bmod p$ in terms of alternating traces of Frobenius on de Rham cohomology.  Then one observes (via the Hodge-to-de Rham spectral sequence) that the trace of Frobenius acting on de Rham cohomology is the same as the trace of Frobenius acting on the coherent cohomology of $\mathcal{O}_X$, essentially because $d(x^p)=0$ in characteristic $p$.  For singular varieties, one needs some slightly more serious argument.
I assume that the proof of Warning's second theorem follows from some Newton-above-Hodge considerations in the case the variety in question is smooth; a cohomological proof seems somewhat more serious in the singular case.
