A better solution than my previous one is

max_{1\leq i \leq n} iN - {i \choose 2}N_2

(That is to say, we can simply consider only i of the sets instead of all n of them, and then apply my previous argument to obtain a lower bound on the size of the union of those i sets, which is also a lower bound on the size of the total union.)

In fact, one can figure out the value of i which maximizes this bound: it will be the largest i for which N - (i-1)N_2 is positive (since this is the difference between the bound obtained using i and that obtained using i-1). This is i = \lfloor N/N_2 \rfloor +1.

So, taking that value for i, the best solution I can see is

nN - {n\choose 2} N_2 if n< i and otherwise iN - {i \choose 2} N_2.

This is, at least, non-negative. In the latter case, it's on the order of N^2/(2N_2).

**

Of course, I'm still not using all the information. I don't see how to get anything further out of inclusion-exclusion when all we have are upper bounds on the sizes of the triple and higher intersections. I encourage anyone who thinks there is an argument in there to post it.

(Edited to correct my arithmetic for the order of the bound in the i <= n case.)