Sure.

By Poonen's Bertini theorem, for every $q$ and $n$, the limit as $d$ goes to $\infty$ of the proportion of degree $d$ hypersurfaces in $\mathbb P^n$ that are smooth and have no $\mathbb F_q$-points is positive.

Being smooth hypersurfaces in $\mathbb P^n$ for (let's say) $n>3$, they have Picard rank $1$, and thus no maps to lower-dimensional varieties.

For an explicit smooth hypersurface of dimension $n-2$, take $\alpha \in \mathbb F_{q^n}$ with $\operatorname{tr}(\alpha)\neq 0$. Choose a basis for $\mathbb F_{q^n}$, so that elements of $\mathbb F_{q^n}$ can be written as vectors over $\mathbb F_q$, with multiplication given by an $n$-tuple of quadratic polynomials.

Consider the function $$ \operatorname{tr} ( \alpha x^{ \frac{q^n-1}{q-1}})$$ from $\mathbb F_{q^n}$ to $\mathbb F_q$. In coordinates, this is a homogeneous polynomial of degree $\frac{q^n-1}{q-1}$ in $n$ variables, hence defines a projective hypersurface, and sends every nonzero element of $\mathbb F_{q^n}$ to a nonzero element of $\mathbb F_q$, hence has no $\mathbb F_q$-points.

Over $\overline{\mathbb F_q}$, we can choose our coordinate system to be the $n$ embeddings of $\mathbb F_{q^n}$ into $\overline{\mathbb F_q}$, and in this coordinate system the equation is $\sum_i c_i x_i^{ \frac{q^n-1}{q-1}}$ where $c_i \neq 0$ is the image of $\alpha_i$ under the $n$th embedding, so this hypersurface is also smooth.

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