For theories with well known prooftheoreticordinals, (what) is there a correspondence between their prooftheoreticordinal and (ordinal indexed families of?) fast growing functions provable total in a given theory ?
Yes: for many theories, $\alpha$ is the prooftheoretic ordinal of T exactly when T proves that $f_\beta$ is total for all $\beta<\alpha$, but does not prove $f_\alpha$ is total. (Where $f_\alpha$ is the $\alpha$th function in the fastgrowing hierarchy.)
Avigad has argued that the correct definition of prooftheoretic ordinals is in terms of provably total functions  specifically, that the proof theoretic ordinal of T is $\geq\alpha$ exactly when T proves the totality of all functions which are "$\prec\alpha$recursive" functions, where $\prec\alpha$recursive means that the function is given by a program together with a timer which uses ordinal notations $\prec\alpha$.
With some care about the encoding, there's a tight connection between proving fastgrowing functions total and proving all $\prec\alpha$recursive functions total, so in some sense one can take proving the totality of fastgrowing functions to be the definition of the prooftheoretic ordinal.
As usual with prooftheoretic ordinals, all reasonable definitions are going to be equivalent for nice theories (which includes all strong enough theories which have appeared naturally elsewhere in logic), but there are artificial theories that make the various definitions no longer equivalent.

$\begingroup$ How come the correspondence you and Avigad are pointing out is not affected by the arbitrariness of choice of fundamental sequences? $\endgroup$ – Łukasz Lew Jun 12 '18 at 0:58

1$\begingroup$ @ŁukaszLew: I think it is somewhat affected by the arbitrariness of choice of fundamental sequences; the fastgrowing hierarchy is less sensitive than the slowgrowing one, but the claim I made is still only going to be for conventional choices of fundamental sequences. (This is one reason Avigad's notion of $\prec$recursive is better as a definition, because some constraints on the encoding are built into the definition, and you can figure out from those which functions grow at the right rates to be provably total.) $\endgroup$ – Henry Towsner Jun 12 '18 at 2:35