A bit more is true when $n=4$. Suppose $F$ is any perfect field, not necessarily finite. Then, for $n=4$, the form $q$ is, up to a scalar multiple, the norm form of a quartic extension of $F$. (Proof: Suppose that $V$ is the projective $F$-scheme defined by $X_1X_2X_3X_4=0$ and that $\overline F$ is an alg. closure of $F$. Then the projective automorphism group of $V$ is a split extension of $T=\mathbb G_m^4$ by the symmetric group $S_4$. Since $H^1(F,T)=0$, the set of isomorphism classes that we seek is $H^1(F,S_4) =Hom(Gal_F,S_4)$, as stated.) For finite $F$ there is only one such extension, of course.
For $n=3$ the configuration of $4$ lines might have triple, or quadruple, points, but if not then again there is only one isomorphism class over $\overline F$. Its automorphism group is $S_4$, and the same argument shows that $q$ is unique up to scalars (you can describe it as a linear section of the quartic norm form.) For $n=2$ I have nothing to add.