Your circuit can be simplified and made pretty easy to quickly guess about:
simulate this circuit– Schematic created using CircuitLab
Note that I've started out dividing in half the current source at the top, half for each side. Tentatively, this flows all the way through the current mirror BJTs, as well.
There's an obvious error to that. The base currents for both \$Q_1\$ and \$Q_2\$ are stolen from \$Q_4\$'s collector current. So \$Q_2\$'s collector current will be about (assuming \$\beta\approx 100\$ as a guess) \$20\:\mu\text{A}\$ less than expected (from \$Q_4\$'s collector.) So \$Q_2\$'s collector current will be a little bit short. This means \$Q_1\$'s collector current will also be short by about that same amount. This leaves \$20\:\mu\text{A}\$ for your \$R_{_\text{OUT}}\$ resistor.
But there's another error now in that thinking. That's because with only \$20\:\mu\text{A}\$ through \$R_{_\text{OUT}}\$, that only makes up \$200\:\text{mV}\$ across \$R_{_\text{OUT}}\$. And that means \$Q_1\$ is being pulled into saturation. This means a little less collector current in \$Q_1\$ and that difference, \$I_x\$, will be added to the current going through \$R_{_\text{OUT}}\$. Perhaps an extra few microamps?
So I'd guess that \$Q_1\$'s collector will be more positive than \$-14.8\:\text{V}\$. But not by a lot. So \$-14.8\:\text{V}\le V_x \le -14.7\:\text{V}\$.
Let's give it a whirl in LTspice:
Hmm. Yeah. Looks close enough for a back of an envelope calculation. I over-estimated the \$V_{_\text{BE}}\$ for the PNP BJTs. But by only \$30\:\text{mV}\$.
I also under-estimated \$\beta\$. But I did that because I was expecting saturation (looking ahead.) If you increase \$R_{_\text{OUT}}\$ by a factor of 10, say, then neither of the current mirror BJTs are saturated (good) and the base currents stolen from \$Q_4\$'s collector current will be closer to the expected amount based on a higher \$\beta\$ value -- perhaps half what I estimated earlier since then \$\beta\approx 200\$, instead. Without saturation for \$Q_1\$, I'd expect the combined two base currents (on the order now of \$2\times 5\:\mu\text{A}=10\:\mu\text{A}\$) to be the current in the new, larger-valued \$R_{_\text{OUT}}\$. So that would be about \$1\:\text{V}\$ drop across \$R_{_\text{OUT}}\$ and now I'd guess \$V_x\approx -14\:\text{V}\$.
Let's see:
Again, fairly predictable with a simple piece of paper and only a few thoughts to put together.