Intro
You write:
I can't seem to figure out how to derive mathematically the values
that I measure in my simulation.
It's easy to suck me into anything mathematical. I admit it. So it'll be a joy to focus on a few initial steps in working out a quantitative answer.
But I already know we'll run quickly into difficulties with the unknown quiescent state of the supposedly class-AB output stage. And that's where I'll need to stop. Still, I want to give you an idea of how to get started. And I'll do this as a stream-of-consciousness and later see if any of what I write can be confirmed, running Spice.
As a segue, the base-emitter voltage of \$Q_3\$ isn't a constant. Many may try to simplify their thinking and analyze the open-loop on that assumption. Assuming that a later Spice run I'm considering confirms the theory I'll apply below now, I shall prove it is not the case and that it's important for any truly quantitative analysis that you also do not assume it is constant.
So let's start.
Early Effect, Without \$Q_3\$ Loading
Identical results should be expected from Spice simulators using identical BJT models in identical situations. I think you already surmised this fact and perhaps this is why you are curious. But your circuit doesn't actually place the 2SC4081 BJTs (\$Q_1\$ and \$Q_2\$) in identical situations. \$Q_1\$ has a resistor in its collector leg; \$Q_2\$ does not. To my mind, this immediately brings up the Early Effect as now the collector voltage will not be the same with both BJTs. (I think you can easily see why.)
So there will already be a small imbalance in collector currents even with identical base voltages. In this case, I'd expect to see \$Q_2\$'s collector current as slightly higher than \$Q_1\$'s collector current (due to how the Early Effect impacts collector currents.)
But by exactly how much? (This isn't immediately obvious, as the following will show.)
Spying at the model found in LTspice for the 2SC4081, I see a model value of \$VA\approx 114\$. So I'd expect a difference of about:
$$100\%-\frac{114\:\text{V}+15\:\text{V}-\left(-620\:\text{mV}\right)}{114\:\text{V}+\left(15\:\text{V}-620\:\text{mV}\right)-\left(-620\:\text{mV}\right)}\approx 0.5\%$$
(Note: I used an assumed \$620\:\text{mV}\$ imagining approximately \$1\:\text{mA}\$ in each collector of the diff-pair. But feel free to vary it a little around that value. You'll see the same percentage arrive. So we can be comfortable with this result.)
This minor imbalance won't impact the base voltage of \$Q_3\$ much. In fact, the impact will be less than initially suggested below, before a negative feedback is applied. But there is a difference. And since there is a difference, let's analyze it. (It's complicated enough to be worth a moment.)
As mentioned parenthetically just a moment ago, assuming about \$1\:\text{mA}\$ in each collector, if perfectly balanced, implies a base-emitter voltage on the 2N5401 BJT (\$Q_3\$) of \$620\:\text{mV}\$.
Without any negative feedback to account for (we'll get to that), the Early Effect alone suggests that the 2N5401 will instead have \$620\:\text{mV}\cdot 99.5\% =616.9\:\text{mV}\$ across its base-emitter junction. That's a change of \$-3.1\:\text{mV}\$ due to the Early Effect when compared to an idealized situation without the Early Effect -- but at this moment still without any loading by \$Q_3\$ taken into account.
Early Effect, With \$Q_3\$ Loading
But there is a negative feedback due to changes in \$Q_3\$'s loading in response to the Early Effect. This was missed above because the computation assumed that \$Q_3\$'s base current didn't load \$Q_1\$'s collector. We know that's not true.
So let's follow the logic to its conclusion. While we are starting with the idea of \$-3.1\:\text{mV}\$ change at the base voltage for \$Q_3\$, there's a negative feedback due to the fact that a reduced \$Q_1\$ collector current implies a reduced base-emitter voltage for \$Q_3\$, which implies a lower base current in \$Q_3\$, which implies less loading on \$Q_1\$'s collector, which finally implies more current returned to \$R_3\$, and therefore more of a voltage drop across \$R_3\$; these results then countering the Early Effect and providing that negative feedback I mentioned.
I won't bore you with the calculus details of the sensitivity equation here, unless you ask for them, but the net effect is that in this case the sensitivity equation gives us about a factor of \$5\$. That factor is applied to the change in \$Q_1\$'s collector current, \$\%\,I_{\text{C}_1}=\frac{\text{d}\,I_{\text{C}_1}}{I_{\text{C}_1}}=-0.5\%\$.
So we should then expect about \$\%\,I_{\text{C}_3}=\frac{\text{d}\,I_{\text{C}_3}}{I_{\text{C}_3}}=5\cdot \frac{\text{d}\,I_{\text{C}_1}}{I_{\text{C}_1}}=5\cdot\%\,I_{\text{C}_1}=-2.5\%\$. This takes into account all of the negative feedback effects due to \$Q_3\$'s loading on \$Q_1\$'s collector current and provides the net \$Q_1\$ and \$Q_2\$ Early Effect on \$Q_3\$'s collector current (ignoring the Early Effect on \$Q_3\$, itself.)
So that's the prediction: differences in the Early Effect between \$Q_1\$ and \$Q_2\$, due to the presence of \$R_3\$, will impact the collector current of \$Q_3\$ by \$-2.5\%\$.
What Does LTspice Say: Confirm or Deny?
Now that I've started you down the path of a quantitative, mathematical analysis, let's see what LTspice actually says about the above calculations I made.
(I did not try LTspice before writing anything up to this point. I'm that confident. But I also make mistakes and so I'm not sure, at all, how things will actually play out in LTspice. So I'm perhaps as curious as you are.)
The idea above is to set up two different models for the 2SC4081. One of them with the nominal value for the Early Effect for the 2SC4081 and another one with the Early Effect neutralized. Then perform two runs and compare their results.
Let's use the captured numbers, shown at bottom in the above capture, and see what we get:
$$\frac{871.05\:\mu\text{A}}{893.336\:\mu\text{A}}\approx 0.975$$
That's exactly the result, as predicted.
Summary
So, I think this shows that you actually can analyze things quantitatively. However, it's non-trivial and requires a variety of skills.
Keep in mind you were attempting an open-loop calculation. Normally, a designer will use (in fact, profoundly rely upon) the use of crafted negative feedback (not just the inadvertent negative feeedback I analyzed above) to cause a discrete circuit to operate within 5-10 percent of the design goals, over the range of operating conditions. So these kinds of detailed calculations I performed above are rarely, if ever, applied.
Now, this is where I stop. I mostly wanted to make the point that you can analyze a circuit and, if you apply enough thinking tools, then you will come credibly close to what Spice will show you. But design efforts are usually not a matter of nailing every decimal point -- leaving that for Spice -- but instead to focus on a bullet-proof design that will work over part and temperature and long term drift variations. Then, Spice is just used to make sure there wasn't some obvious mistake and also to verify that the rough estimates are in the right ballpark.
