Well, first of all you're making a mistake re: the slow-growing hierarchy - it's not indexed by ordinals, but rather families of fundamental sequences of ordinals (or copies of ordinals, or etc.). This is an unavoidable difference, since there is in general no good way to assign a fundamental sequence (or copy) to every countable ordinal.
Incidentally, this can be made precise in various ways - for example, it's consistent with ZF (= set theory without the axiom of choice) that there is no function assigning each countable ordinal a fundamental sequence at all, and "lower down" there are computability-theoretic obstacles to assigning fundamental sequences to each computable ordinal simultaneously. (And this is really a general point about hierarchies through the countable ordinals, not the slow-growing hierarchy specifically.)
The slow- or fast-growing hierarchies we care about are for this reason not defined through all the countable ordinals, but rather only up to some fixed countable ordinal leading up to which we've already defined a family of fundamental sequences ($\epsilon_0$ is a common stopping point). I've certainly never seen such a hierarchy continued even through all the computable ordinals, and $\alpha$ is far, far greater than any computable ordinal.
Having said that, the answer to your question is yes. For any countable ordinal $\mu$ whatsoever, there is an assignment $\mathfrak{S}$ of fundamental sequences to every limit ordinal $\le\mu$ (this $\mathfrak{S}$ wont' be unique, but such a thing will exist). Using any such $\mathfrak{S}$ we can assign a slow-growing function to each ordinal $\le\mu$ in the usual way. But note that the dependence on the specific choice of $\mathfrak{S}$, coupled with the lack of a general way to pick a specific $\mathfrak{S}$, means that in general there is no specific function deserving to be called "the $\mu$th slow-growing function."
In particular, the special properties of $\alpha$ are completely irrelevant here. Any countable ordinal (including $\alpha$ if it exists) has fundamental-sequence-families,$^*$ and with respect to each such family there will be an $\alpha$th corresponding slow-growing function.
$^*$Why is this? Well, perhaps surprisingly in light of the weird ZF fact mentioned above, this has an easy proof! And one which makes use of the notion of a copy of an ordinal.
Specifically, suppose $\theta$ is a countable limit ordinal. Since it's countable, there's a well-ordering $R$ of $\omega$ with order-type $\theta$. (Note that this $R$ isn't unique - and in fact there's no way to "canonically" pick $R$, and this is why this proof doesn't contradict the weird ZF fact mentioned above!) I'm going to use $R$ to get a fundamental sequence for every limit ordinal $\le \theta$.
Each $n\in\omega$ has, via $R$, a corresponding ordinal $[n]_R<\theta$: namely, the ordinal corresponding to the ordertype of the set $\{m\in\omega: mRn\}$ ordered by (the restriction of) $R$. For example, taking $\theta=\omega+\omega$ and $R$ to be the ordering $$1\prec 3\prec 5\prec 7\prec ...\prec 0\prec 2\prec 4\prec 6\prec ...,$$ we have $[6]_R=\omega+3$. Now for $\eta\le\theta$ a limit:
We first define a sequence $(n^\eta_i)_{i\in\omega}$ of natural numbers given by $$n^\eta_i=\min\{m\in\omega: [m]_R<\eta, m>n^\eta_j\mbox{ for all $j<i$}\}.$$ Importantly, the first "$<$" there refers to the usual ordering on the ordinals, while the "$\min$," the "$>$," and the second "$<$" refer to the usual ordering on the naturals.
We now use this to define a sequence of ordinals $(\gamma_i^\eta)_{i\in\omega}$ of ordinals $\le\eta$ given by $\gamma^\eta_i=([n^\eta_i]_R)_{i\in\omega}$.
It's now easy to check that for each limit ordinal $\eta\le\theta$ the sequence $(\gamma_i^\eta)_{i\in\omega}$ is a fundamental sequence for $\eta$.
For example, taking $R$ as above and $\eta=\theta=\omega+\omega$, we get $$(n_i^\eta)_{i\in\omega}=(1,2,4,6,8,10,...),$$ and this gives the fundamental sequence $$0,\omega+1,\omega+2,\omega+3,...$$
(because $[1]_R=0,[2]_R=\omega+1,[4]_R=\omega+2$, ...).
