Well, J_0(p) itself is known to have *analytic* rank at most (6.5) dim J_0(p), by a result of Kowalski and Michel:

http://www.math.u-bordeaux1.fr/~kowalski/explicit-rank.pdf

so that's an upper bound to start with, at least if you are willing to condition on BSD. An unconditional upper bound on the rank seems hard -- I'm not sure how you would hope to rule out the existence of (e.g.) some elliptic curve of conductor p whose Mordell-Weil rank is gigantic.

The dimension of the Zariski closure is just the sum of all the A_f with positive Mordell-Weil rank, right? Here you might be able to make slightly more progress. What you are looking for is an lower bound on the number of cuspforms f in S_2(Gamma_0(p)) such that L(f,1) is nonzero. (This is unconditional, since we know that nonvanishing implies rank 0.) Of course, it's not even obvious that there is one such cuspform! One way to show the existence is to compute a suitable weighted average of the L(f,1) via Petersson formula. The average will be nonzero, which means one of the special values is nonzero. By being more clever about the many choices of average you can take, you can get lower bounds; I think Amir Akbary knows a lot about this.

But you may want something much stronger; e.g., that the sum of dim A_f over all f with L(f,1)=0 is o(dim J_0(p)). That's what one would expect. I don't know that anyone has proved anything of this kind.