The answer is yes. A Levi subgroup is connected reductive, and hence the semisimple elements are dense. Each ss elt lives inside a max torus. So the Levi is generated by its maximal tori, each of which is a maximal torus of G.
– Peter McNamaraMay 9 '12 at 16:47

Thanks Peter for the correction, I misread "generated by maximal tori" by "generated by one maximal torus". I deleted my wrong example. Nevertheless, this is better suited for math.exchange.
– Marc PalmMay 9 '12 at 17:05

I agree that this should be asked in a less research-oriented forum, since it's an easy result of the standard development of Borel-Chevalley structure theory. On the other hand, it's a good exercise in combining the density of semisimple elements in a connected reductive group and the fact that each lies in a maximal torus with the elementary fact that all tori generate a closed connected subgroup of the given group. But generation by all maximal tori is seldom if ever used in practice. The detailed Bruhat decomposition along with study of root groups are more essential.
– Jim HumphreysMay 9 '12 at 18:54