It sounds like a random question, and it is, but its been bugging me, if you either start with $2$ or $e$ - it comes to the same thing - which is an arbitrary starting point that I've chosen, what is the lowest number in can generate, can it go infinitely low? I don't expect anyone to be able to give me an answer but if anyone knows a good way to approach this I'd appreciate it.
Background: I was messing around on my calculator when I stumbled upon $\ln(\mathrm{ans}^2)$, when I iterated from $2$, it didn't grow exponentially or tend towards a value, as most iterations do, instead it jumped around, positive and negative, never getting very large. Every now and then it would spike. I wrote a C++ program to find lower numbers, but the lowest it could find was around $-36$, I say around not because I can't remember, but because the different platforms I ran my code on gave different answers, presumably due to small rounding differences. In addition, $-36$ doesn't come up quickly either, it's over a billion iterations in before it appears, and then nothing. I tried to solve $x = \ln(x^2)$ to find a value that would return itself but couldn't, by trial and error I found that $-0.7$ (and a lot more decimals) was close, but drifted away after a few iterations. A little later I also discovered that $-\sqrt{e^x}$ tended towards this value.
Speculation: it seems plausible that somewhere after $-36$, it gets stuck in a loop, but that loop could be well over a million numbers long, so I can't know. If it does get arbitrarily low, is there a formula for where a certain lowness will be achieved, are all numbers generated eventually? I have no idea. Does anyone?
(Partial Answer)
Firstly, $x = \ln(x^2)$ is impossible for positive $x$. This is because $\ln$ is convex, which means that any tangent at $(x_0, \ln(x_0^2))$ will be an upper bound of $\ln(x^2)$ for all $x$. In particular, at $(2, \ln(4))$ we get the tangent line $y = x - 2 + 2\ln(2)$ i.e. $x - 2 + 2\ln(2) \geq \ln(x^2)$ for all $x$ (with equality at $x = 2$ only), and $y = x$ lies "above" this line.
However, it is possible for $x < 0$. Indeed, zooming to the left gives you the full picture, where $x_0 \approx -0.703$ is a fixed point of $x \mapsto \ln(x^2)$. Apart from the numerical value, we can actually say more about the behaviour of the map around this fixed point. For example, what if we don't have the exact value, but only an approximation, say $x_0 = -0.7$ as you tried on your calculator? Well if you have tried this for other iterative maps before, say $x \mapsto \sqrt{|x|}$, you will usually still converge to a fixed point simply by iterating the function over and over again. However, that's not the case here! The reason is that the fixed point is unstable. You can see this by looking at the derivative or the slope of the curve around the fixed point, and there are resources to read on this interesting topic.
For your question, we can formulate it mathematically using a sequence. Define the sequence $a_0 = 2$ and $a_{n + 1} = f(a_n) := \ln(a_n^2) = 2\ln(|a_n|)$, and we want to see whether $a_n$ is bounded below. Firstly, I can confirm that the $-36$ result should be achievable, I got $x_{5644271} \approx -37.4046$.
The problem is related to a field of mathematics called dynamical systems, which studies the behaviour of iterating these maps, and the problem seems to be hard, so I don't have an answer. However, one idea you can try is to reverse the map. In particular, say you want a particular $a_k$ to be very small in size, since that means the next step would be "very negative". We can write this as $a_k \in (-\epsilon, 0) \cup (0, \epsilon)$ (excluding $0$). Then, we consider the inverse function, which is given by $f^{-1}(a) = \{e^{a / 2}, -e^{a / 2}\}$. This means the preimages $f^{-1}((-\epsilon, 0)) = (\exp(-\epsilon / 2), 1)$ with its negative. Now, we might approximate $\exp(x)$ by its taylor series $\exp(x) = 1 + x + \frac{x^2}{2} + \cdots$, so the interval is approximately $(1 - \frac{\epsilon}{2}, 1)$. I am running into precision issues extremely quickly with my code, but the idea is if you can prove that this inverse map will "enlarge" the size of the interval, then by iterating the inverse map you will get that every starting value $x$ must reach $(-\epsilon, 0) \cup (0, \epsilon)$ at some point.
Hope this helps a bit, cheers!