The trivial bound $m=q+1$ mentioned in the case where $n=2$ is actually true for all $n$ by simple counting: a set of $k$ hyperplanes contains at most $kq^{n-1}+q^{n-2}+\ldots+1$ points, since the "first" hyperplane has $q^{n-1}+\ldots+1$ points, then each subsequent hyperplane meets the "first" hyperplane in $q^{n-2}+\ldots+1$ points, and thus adds at most $q^{n-1}$ new points to the union. But this bound clearly cannot be met without violating the condition that the intersection of all hyperplanes be trivial.
In the plane ($n=2$), the actual lower bound on a blocking set is $q+\sqrt{q}+1$, due to Bruen [Bruen, A., Blocking sets finite projective planes, SIAM J. Appl. Math. 21 (1971), 380–892]. The lower bound is met if and only if the blocking set is a Baer subplane (i.e., a subplane of order $\sqrt{q}$), which occurs only if $q$ is square. There is a substantial literature regarding the situation when $q$ is not square; Hirschfeld [Hirschfeld, J W P, 'Blocking sets', Projective Geometries over Finite Fields (Oxford, 1998; online edn, Oxford Academic, 31 Oct. 2023), https://doi.org/10.1093/oso/9780198502951.003.0013, accessed 10 Nov. 2024.] provides an entire chapter surveying results in this direction.
The situation in higher dimensions is not nearly as well studied. A good reference which addresses some of the issues is Sziklai [P. Sziklai. On small blocking sets and their linearity. J. Combin. Theory
Ser. A, 115:1167–1182, 2008.]. In particular, this article contains some relatively small examples of blocking sets which are based on subgeometries, but I do not believe there is any sort of general proof of a lower bound here.
