Today, I heard that people think that if you can prove the Hodge conjecture for abelian varieties, then it should be true in general.  Apparently this case is important enough (and hard enough) that Weil wrote up some families of abelian 4-folds that were potential counterexamples to the Hodge conjecture, but I've never heard of another potential counterexample.

Anyway, in short:

1) Does the Hodge Conjecture for abelian varieties imply the full Hodge conjecture?

2) If not, is there an intuitive reason why abelian varieties should be the hardest case?

I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties of CM-type, and we can't even prove that the Hodge classes Weil described are algebraic. So if the Hodge conjecture was proved in one interesting case, e.g., abelian varieties, that would be a big boost.

Added: As follow up to Matt Emerton's answer, a proof that the Hodge conjecture for abelian varieties implies the Hodge conjecture for all varieties would (surely) also show that Deligne's theorem (that Hodge classes on abelian varieties are absolutely Hodge) implies the same statement for all varieties. But no such result is known (and would be extremely interesting).
