Rhene Thom put the same question to Sullivan duaring Sullivan's talk at IHES in 1980: "Why Lie groups are null-cobordant?" Sullivan answered: "It follows from Thom's theorem" (A manifold is null-cobordant precisely when each of its characteristic numbers vanish.)
Thom: "OK, but can you prove this without this heavy tool?"
After few minutes Sullivan said:
"Take a circle subgroup in the Lie group. Take all the posets. This gives a fibration with fiber G (the Lie group) and fiber S^1. Take the associted disc bundle. Its boundary is G." Q.E.D.
An elementary proof that paralellizable manifold is null-cobordant can be found in a paper by Buoncristiano- Hacon. I am not sure it gives oriented null-cobordism.