A recent article of Bhargava and Shankar, "Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves" (http://arxiv.org/abs/1006.1002), addresses, among many other related questions, the density of monogenic cubic orders, counted by the height (slightly modified) of their invariants $I$, $J$.

This height is almost the same as the height of the discriminant of the polynomial to which they are associated with. So this answers the first case above for cubic fields.

Theorem 4.1 (Bhargava, Shankar):

Let $N_3^{(0)}(X,\delta)$ (resp. $N_3^{(1)}(X,\delta)$) denote
the number of cubic submonogenized rings $(C,x)$ of index $n$ with
positive (resp. negative) discriminant such that $H(C,x)\lt X$ and
$n\lt X^\delta$, where $\delta \leq 1/4$. Then we have
$$N_3^{(0)}(X,\delta)=\displaystyle\frac{4}{45}X^{5/6\,+\,2\delta/3}+O(X^{5/6});$$
$$N_3^{(1)}(X,\delta)=\displaystyle\frac{16}{45}X^{5/6\,+\,2\delta/3}+O(X^{5/6}).$$

If I understand correctly, this means that the density for case (1) above (restricted to cubic fields) is in fact 0.

The article goes on to study quartic rings, but the result, Theorem4.8, is a bit more difficult to understand. But if I do, it says (a bit more than) that **the density of monogenic quartic fields counted by discriminant is, once more, 0**. Note that in the theorem the count is by the invariants of resolvent, which is like the discriminant of the resolvent, which is a constant factor away from the discriminant of the quartic polynomial to which it is associated with.

So, an educated guess is that the answer to (1) above is 0 for all degrees, and then in general (as stated above without degree). On the other hand, the discriminant of an $n$ degree polynomial is a multivariate polynomial, and I suspect that sieve theory can prove that for every constant degree there is a positive density (counted as in (2)) of polynomials that have discriminant with bounded squareful part. This in turn would prove that there is a positive density of polynomials giving orders of bounded index in their respective maximal orders. Taking this educated guess a little further, I suspect the answer to (2) is greater than 0, contrary to (1).