This question is based on the assumption that all computations start with no ordinal parameters (i.e. the input is empty).
The term “stabilization time of a machine” for this question implies the successor of $\alpha$, where $\alpha$ is the least ordinal such that the values of all symbols written on all cells of the initial segment of length $\omega$ of the output tape never change at any time $\beta > \alpha$. If a machine diverges (i.e. the initial segment of length $\omega$ of the output tape does not stabilize), then the stabilization time of such machine is $0$.
Consider Ordinal Turing Machines with no oracle. Let us call such machines “$0$-th order machines”. There exists the supremum of stabilization times (let us denote it by $\sigma(0)$) of these machines.
Consider Ordinal Turing Machines with the oracle tape. This additional tape has a single cell (namely, the $\sigma(0)$-th one) marked by a non-zero symbol. Let us call such machines “$1$-st order machines”. There exists the supremum of stabilization times (let us denote it by $\sigma(1)$) of these machines.
Then consider Ordinal Turing Machines with the oracle tape where two cells (namely, the $\sigma(0)$-th and $\sigma(1)$-th ones) are marked by non-zero symbols. Let us call such machines “$2$-nd order machines”. There exists the supremum of stabilization times (let us denote it by $\sigma(2)$) of these machines.
Similarly, one can define the ordinals $\sigma(n)$ for any natural number $n$.
The oracle tape of the $\omega$-th order machines contains $\omega$ many cells marked by non-zero symbols: a particular cell corresponds to some $\sigma(n)$. There exists the supremum of stabilization times (let us denote it by $\sigma(\omega)$) of these machines.
We can continue: the oracle tape of the $(\omega+1)$-th order machines contains $\omega+1$ many cells marked by non-zero symbols: a particular cell corresponds to some $\sigma(\alpha)$, where $\alpha < \omega+1$. Note that the $\sigma(\omega)$-th cell is the last cell of the oracle tape marked by a non-zero symbol. There exists the supremum of stabilization times (let us denote it by $\sigma(\omega+1)$) of these machines.
Can we continue the similar logic unboundedly and define the class $\mathsf{S}$ of ordinals of the form $\sigma(\beta)$, where $\beta$ may be an arbitrarily large (possibly uncountable) ordinal?
If no, why?
If yes, consider Ordinal Turing Machines with the oracle tape where an $\alpha$-th cell is marked by a non-zero symbol if and only if $\alpha$ belongs to $\mathsf{S}$ (i.e. if and only if $\alpha = \sigma(\beta)$ for some ordinal $\beta$). Let us call such machines “$\mathsf{S}$-machines”.
Let $\tau(0)$ denote the supremum of countable ordinals eventually writable by $\mathsf{S}$-machines on the initial segment of length $\omega$ of the output tape.
Let $\mu(3)$ denote the least ordinal $\delta$ such that $L_{\delta} \prec_{\Sigma_3} L_{\omega_1}$. This ordinal is mentioned in this comment and this answer on Mathoverflow.
Question: which ordinal is larger, $\tau(0)$ or $\mu(3)$?
(I do not know whether the answers depend on the acceptance of the axiom of constructibility, but if they do, I am interested in both cases.)
