We claim it is possible iff $n\equiv 0, 2\pmod{3}$. Notice that an $n$-level honeycomb can be tiled by three $n \times n+1$ rhombi, arranged in a rotationally symmetrical way around the middle hexagon. If $n\equiv 0, 2\pmod{3}$, then $n$ or $n+1$ respectively are divisible by 3, so each rhombus can be tiled by trexes. This shows it is possible in these cases. Now, suppose $n\equiv 1\pmod{3}$ and $n>1$ (it is clearly impossible when $n=1$). Rotate the honeycomb so that one of its vertices is pointing vertically downwards. Label the rows of hexagons $1$ to $4n+1$, and colour the hexagons in the rows congruent to $0, 1, 2\pmod{3}$ red, green, blue respectively. Notice that every trex covers exactly one of each colour hexagon, so if a tiling is possible then each colour appears the same number of times. However it can easily be seen that this is not the case in each of the aforementioned rhombi as neither of the side lengths is divisible by 3, and as each rhombus is coloured identically we reach the conclusion that no such tiling exists.

@above I think you meant $n+1$?

Yeah thanks