My implicit assumption was by Faltings height. It seems not improbable, though, that the ordering them by conductor would give the same average.[?]

Indeed they do, and it is not at all clear what the answer should be.

Thank you very much for those references. The bounds for the hyperelliptic Jacobians in the Bhargava-Gross paper being the same as for elliptic curves, it could perhaps not be too wild to believe that (contrary to what I expected) the same distribution persists for abelian varieties: for a given global field $F$ and a given dimension $g$, that $50$% of the $g$-dimensional abelian varieties over $F$ have $0$ rank, $50$% have rank $1$, and $0$% have higher rank?

@ulrich: Thanks for the example! @ACL: I had in mind proper varieties.

Yes! So, in char. 0, $X$ is rationally connected iff. it has a smooth projective rationally connected (birational) model. But since resolution of singularities is not available in positive characteristic, this definition is the general one. In general, in any characteristic, a smooth separably rationally connected variety is simply connected. (And as noted above, all we can say about non-separably rationally connected varieties in char. $p > 0$ is that their fundamental group is finite of order prime to $p$).

A normal projective variety $X$ is (separably) rationally connected if there exists a finite-type integral scheme $T$ and a morphism $F : T \times \mathbb{P}^1 \to X$ such that the obvious morphism $T \times \mathbb{P}^1 \times \mathbb{P}^1 \to X \times X$ is dominant (and generically smooth).

I had the same point as Dmitri. In expose X it is shown, as an application of the Zariski-Nagata purity theorem, that the algebraic fundamental group of a complete regular variety is birationally invariant. In particular, this certainly contains a proof of the simple connectedness of regular proper rational varieties over an alg. closed field. Statement XI, Cor. 1.2 simply refers to this birational invariance. Does the birational invariance extend to normal varieties?

... where the latter, by the Lefschetz formula, is also equal, simply, to $\prod_{x}' (1-|k(x)^{-s}|)^{-1}$, where the product is over the closed points $x$ of $X$ of residue characteristic not in $S$.

If you take a smooth model $X$ over the localization $\mathbb{Z}_S$ at a finite set of primes $S$ (which is always possible by clearing denominators), then your partial zeta function is $\prod_i \prod_{p \notin S} \det \Big( 1 - p^{-s}\mathrm{Frob}_p | H^i(\overline{X}_{\mathbb{Z}/p},\mathbb{Q}_l) \Big)^{(-1)^i}$.

Yes, I did not intend to write that there is no $\mathbb{Z}$-model in general, only that there is no canonical choice.

Yes, there is no preferred integral model in general (with the notable exception of curves and abelian varieties). If you take an arbitrary integral model, you would only get the zeta function up an undetermined rational factor, which for many purposes is fine. But for obtaining the precise zeta function, the way out is to look at the Frobenius action on $H^i(\overline{X},\mathbb{Q}_{l})^{I_p}$ (which is well-defined precisely because we take inertial invariants).