Smooth complete intersections and sharpness of the Chevalley-Warning theorem Let $X$ be a complete intersection in $\mathbb{P}^n$ of multidegree $(d_1,\ldots,d_r)$. If we're working over a finite field $\mathbb{F}_q$, the Ax-Chevalley-Warning theorem says that if $X$ is in the Fano range, i.e., $$ \sum d_i \leq n,$$ then $|X(\mathbb{F}_q)| \equiv 1 \pmod q.$
In the case that $X$ is not in the Fano range, one can cook up examples of such $X$ with no $\mathbb{F}_q$-points at all using the norm form, but these are not smooth.
In SGA 7 II Exposè XXI, Katz shows that for a general complete intersection in the Fano range, this congruence is not satisfied, possibly after extending the ground field.
Given a multidegree outside the Fano range, are there explicit examples of smooth complete intersections which don't satisfy this congruence?
Edit: Sorry, I wasn't sufficiently explicit about what I was asking. Restricting to hypersurfaces, what I would like is for each $d$, $p$, and $n$ with $d \geq n+1$, an explicit example of a smooth hypersurface of degree $d$ in $\mathbb{P}^n$ which doesn't satisfy the Chevalley-Warning congruence.
Thanks for the examples, though; I knew some explicit examples before but no infinite families.
 A: Extension of Daniel Loughan's example (which is also known, but not as
well-known as it should(?) be):  if a prime $p$ is of the form $dn+1$ then
the Fermat hypersurface $\sum_{i=1}^d x_i^d = 0$ in ${\bf P}^{d-1}({\bf F}_q)$
is smooth and its number of rational points is not congruent to $1 \bmod p$. 
Indeed the usual argument for Chevalley(-Warning) shows that
the number of rational points is congruent mod $p$ to $1 \pm t$
where $t$ is the $(x_1 x_2 \cdots x_d)^{p-1}$ coefficient of
$\left(\sum_{i=1}^d x_i^d\right)^{p-1}$, and when $p = dn+1$
this coefficient is $(p-1)!/n!^d$ which is clearly not $0 \bmod p$. 
A: How about $$x_1^{q-1} + \cdots + x_{q-1}^{q-1} = 0  \subset \mathbb{P}^{q-2} \quad ?$$
This is a fairly well-known example which seems to satisfy your criteria.
A: How explicit is "explicit"?  Even norms of field extensions are somewhat inexplicit.  
The following gives just one example, but the method extends to other examples.  If you have a pencil of degree $d$ hypersurfaces in $\mathbb{P}^n$ over $\mathbb{F}_q$, if you can count the number of points in the base locus, and if you can count the number of (rational) singular members of the pencil and the number of points of each singular fiber, then you can sometimes prove that the number of points in some (rational) smooth member of the pencil cannot be congruent to $1$ modulo $q$.  For instance, if $q=2^r$ for $r\geq 2$, this applies to the pencil from the following: When is the kernel of the etale fundamental group in a fibration abelian? 
That example is a pencil of degree $4$ plane curves, the base locus has $4$ rational points, there are precisely two singular fibers, both rational, one of which has $4q-2 = (4q-6)+4$ rational points, and the second of which has $q+1 = (q-3)+4$ rational points.  The union of the singular fibers has $4+(4q-6)+(q-3) = 5q-5$ rational points.  Thus, the number of rational points contained in no singular fiber is $$(q^2+q+1)-(5q-5) = q^2-4q+6.$$ There are $q-1$ smooth fibers.  If the number of points in a smooth fiber is congruent to $1$ modulo $q$, then the number of non-base points is congruent to $1-4=-3$ modulo $q$.  Modulo $q$, the total number of non-base points in smooth fibers is congruent to $$-3(q-1) = -3q+3.$$  Since $q=2^r$, $6$ is not congruent to $3$ modulo $q$.  Thus, at least one smooth fiber violates the congruence.  
Edit. This method seems to apply easily to other pencils.  For instance, for every $q$ it follows that for some $a\in \mathbb{F}_q^\times$, the plane cubic curve $\text{Zero}(z^3-axy(y-x))$ in $\mathbb{P}^2$ violates the congruence.
