= Solution
Every nonzero vector can be extended to a <basis>, and the invertible map between two such bases carries the first vector to the second. The action is therefore transitive. Counting ordered bases gives
$$
|G|=(8-1)(8-2)(8-4)=168.
$$
The <stabilizer> $H$ of one nonzero vector consequently has order $168/7=24$. There are three planes containing that vector, corresponding to the three lines of the two-dimensional quotient space. Adapted bases show $H$ acts transitively on these planes, so the <stabilizer> of one such plane has order $24/3=8$.
Let $Q$ be any subgroup of order eight. Its <orbit> lengths on the seven nonzero vectors are powers of two. Since their sum is odd, at least one <orbit> has length one. Fix such a vector $v$. The action on the three planes through $v$ likewise has a fixed plane $P$. Thus $Q$ is contained in the <stabilizer> of the <flag> $\langle v\rangle\subset P$, and equality follows because that <stabilizer> already has order eight. The full <group> acts transitively on these flags, so all the subgroups are conjugate.
There are $7\cdot3=21$ flags. To see they yield distinct subgroups, use the standard upper unitriangular matrices relative to an adapted <basis>. Their only common nonzero fixed vector is the first <basis> vector: varying entries in positions $(1,2)$ and $(1,3)$ forces the other two coordinates to vanish. On the quotient by that line, varying position $(2,3)$ leaves only the line of the second <basis> vector fixed. Hence this subgroup fixes exactly one full <flag>. Conjugation transfers uniqueness to each <flag> <stabilizer>, giving
$$
\boxed{21\text{ subgroups of order eight}}.
$$
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