For all groups of exceptional types the answer can be deduced from the paper "Jordan block sizes of unipotent elements in exceptional algebraic groups" by Ross Lawther, published in Comm. Algebra, 23, Issue 11, 1995, 4125-4156. In order to determine the Jordan block sizes of ${\rm Ad}\ u$, where $u\in G$ is unipotent, Lawther used comuter-aided computations. The case of a general $x\in G$ reduces quickly to the case where $x$ is unipotent by using the Jordan-Chevalley decomposition in $G$.
Of course, the number of Jordan blocks is just $\dim\ \mathfrak{c}_{\mathfrak g}(u)$, or rather $\dim\ \mathfrak{g}^{{\rm Ad}\ u}$, which is one of the numbers you are interested in. This number could then be compared with $\dim\ C_G(u)$ which is much easier to find in the literature (Spaltenstein's book on unipotent classes might have an answer).
Regarding Steinberg's Problem 12, when $p=2$ the Lie lagebra of type $E_8$ is simple, hence has zero centre, but there are about 20 instances when the value of $\dim\ \mathfrak{g}^{{\rm Ad}\ u}$ jumps and becomes bigger than that for the counterpart of $u$ in good characteistic (these cases are starred in Lawther's paper). Since the number of "new" unipotent classes in characteristic $2$ is far less than 20, I think, there will be instances when $\dim\ \mathfrak{g}^{{\rm Ad}\ u}>\dim\ C_G(u)$. The orbit labelled $E_7$ could do the trick, but I didn't check.
There is also a version of Lawther's results for nilpotent elements in exceptional Lie lagebras in the literature; see "Varieties of nilpotent elements for simple Lie algebras II: Bad primes" by University of Georgia VIGRE Algebra Group, published in J. Algebra, 292, 2005, 65–99. This paper also relies on extensive computer-aided computations.
As fas far as I'm aware there is no computer-free proof of the above results at the present time, hence there is no uniform proof either. One could even say, on a more philosophical note, that nothing is uniform when $p$ is bad.
