$(x_1-x_2-x_3+x_4)(x_2-x_4-x_3+x_1)(x_3-x_1)(x_4-x_2)\equiv 0$ over $\mathbb F_3$.

Take polynomials $p_1,\dots,p_n$ over variables $x_1,\dots,x_n$ such that $p_i$ does *not* depend on $x_i$ and $\Pi_{i=1}^n (x_i+p_i) \equiv 0$ over $\mathbb F_q$.  
It is easy to see that one can take $n=q$ and $p_i=i+\sum_{j\ne i} x_j$ to satisfy this equation.  
If each $p_i$ can depend on only a bounded number of variables, then it follows from the Local Lemma that this is not possible.  
It is not that hard to show that for $n=2^{q-1}$ there are polynomials such that each $p_i$ depends on only one $x_j$ with $j<i$; see the example in the first line for $q=3$.  

In general, have such polynomials been studied in number theory?

My motivation comes from that this is an equivalent reformulation of the [Hat Guessing Number][1] of graphs.  
To decide whether such polynimials exists such that their [dependency graph is degenerate][2], would solve an interesting open problem. This graph is defined on $n$ vertices such that $ij\in E$ if $p_i$ depends on $x_j$ or $p_j$ depends on $x_i$. A graph is $d$-degenerate if its vertices can be ordered such that from each vertex $v_i$ there are at most $d$ edges to vertices coming after $v_i$ in this ordering. In our problem $d$ can be any constant, compared to which $q$ and $n$ can be arbitrary.


  [1]: https://arxiv.org/abs/1812.09752
  [2]: https://arxiv.org/abs/2003.04990