It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow  {\mathbb F}_q$ is the map $x\mapsto x^p$. 

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$.
The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra  $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$. 

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].