One can learn that for a power tower of height $n$:
$$y=n\big\{x^{x^{x^…}}=\,^nx\implies \log_x(y)=\boxed{\log_x(\,^nx)=\,^{n-1}x}$$
giving a recursive relation. One might see that $n<-2$ cases are undefined, but using the extended real numbers or limits, one can actually use the principal branch for the logarithm function for $x\ne0$:
$$\log_x(\,^1x)=\frac{\ln(\,^1 x)}{\ln(x)}=\,^{1-1}x=\,^0x=1\implies \frac{\ln(1)}{\ln(x)}=\,^{0-1}x=\,^{-1}x =0$$
Here is when problems occur. We can define $\ln(-\infty)=\infty$ or $\ln(-\infty)=\ln(-1)+\ln(\infty)=\pi i+\infty$, but to stick with real values, like in
let us consider the simpler version of $\boxed{\ln(-\infty)=\infty}$. Note that I will not simplify the logarithms into the infinity to keep the domain as large as possible:
$$\frac{\ln(\,^{-1} x)}{\ln(x)}=\frac{\ln(0)}{\ln(x)}=\,^{{-1}-1}x=\,^{-2}x =-\frac{\infty}{\ln(x)}\\\implies \frac{\ln(\,^{-2} x)}{\ln(x)}= \frac{\ln\left(\frac{-\infty}{\ln(x)}\right)}{\ln(x)}=\,^{{-2}-1}x=\,^{-3}x = \frac{\ln\left(\frac{-\infty}{\ln(x)}\right)}{\ln(x)} = \frac{\infty-\ln(\ln(x))}{\ln(x)}\\\implies \frac{\ln(\,^{-3} x)}{\ln(x)}=\frac{\ln\left(\frac{\infty-\ln(\ln(x))}{\ln(x)}\right)}{\ln(x)}=\,^{{-3}-1}x=\,^{-4}x = \frac{\ln\left(\frac{\infty-\ln(\ln(x))}{\ln(x)}\right)}{\ln(x)}=\log_x\left(\infty-\ln(\ln(x))\right)-\log_x(\ln(x)) $$
and so on. This creates a couple questions using our aforementioned assumptions and these Wolfram Functions conventions. The question here is what would happen if we calculated:
$$\lim_{n\to-\infty}\,^n x=\lim_{n\to\infty}\underbrace{\log_x\left(…\log_x(0)\right)}_n=\,^{-\infty} x$$ which looks like a W-Lambert function: $$-\frac{\text W_{-1}(-\ln(x))}{\ln(x)}= \lim_{n\to\infty}\underbrace{\log_x\left(…\log_x(e)\right)}_n $$ using the same process and the recursive definition of:
$$\log_x(\,^nx)=\frac{\ln(\,^nx)}{\ln(x)}=\,^{n-1}x $$ Also, what is the range of this new experimental $\ \,^{-\infty} x\ $ function? I have listed the conventions that will be used, so there should be no confusion. Please correct me and give me feedback!
(Not an answer, only an extension of my 2.nd comment)
I show some iterates of the $\log()$ with base=$e$. Using my proposed definition for the notation we see some initial values $z_0$ between $0$ and $1$ and their $h$'th iterates $z_h$. I think it is easily extrapolatable, where the $h$'th iterates tend to, when $z_0$ approaches either the initial value $0$ (or $1$, which is implied by your notation). This is a q&d-computation with Pari/GP with internal precision of 200 dec digits.
Below there is a plot of the iterates $z_h = \log°^h(z_0)$ of the line $z_0 \in \varepsilon \ldots 1-\varepsilon$
The lines $z_0 , z_1=\log°^1(z_0),z_2 =\log°^2(z_0)$ are not displayed. The line for $z_3$ is the red curve in the right half looking roughly like a $\sqrt x $ curve, the line for $z_4$ is the green curve most neighboured at the $z_3$ curve and the positive real axis, whose both ends tend to $+ \infty + 0 \pi î$.
This plot should work as a suggestion how -if at all- that infinite iteration could be defined most meaningfully.
Additional remark: the curves could not have been drawn out to the values $x \approx 9$ by the iteration of logarithms alone; two times iterated back we had a real value of $\exp(\exp(9.4)) \approx 2.1 E 4750$ which gives then a limit for $z_0$ near $0$ which I wouldn't have used from the beginning . But there is a functional relation between the real- and the imaginary value in the $z_3$ curve which allows to compute (and plot) the curve to arbitrary real-argument larger than this roughly 1.2-value. From that the iterated logarithms for a far wider interval of the $z_3$ can be computed. (Sorry, don't have that functional relation at hand at the moment, but was simply derived from the trigonometric formula for the complex logarithm).