You can calculate Vo by making the unrealistic hypothesis that all transistors are perfectly identical. This will not be the case using real transistors though, so this calculation is useless.
Q1 and Q2 form a long tail pair differential amplifier. They take as input the voltage difference between their bases, and output a difference in collector currents. It is important to notice that the output is a current, not a voltage.
Here, both bases are at the same potential. Using the unrealistic hypothesis that both transistors are identical, their collector currents are thus equal, and their value is I1/2 minus the base current, which we will neglect.
Now, Q3 and Q4 form a current mirror. Q1's collector current is mirrored and output by Q3. Q3's collector current is thus I1/2 also.
Q3 and Q1's collector currents thus cancel perfectly and... we forgot Q3's base current. We can't neglect it this time.
Thus, after the I1/2 parts in Q1 and Q3 collector currents cancel, the current which attempts to flow into node "Vo" is the base current of Q3, which is negative.
With I1=1mA and Q3 having a hFe of 200, this current is 1mA/2/200 = 2.5µA
However, we neglected Early effect (and self heating) in all transistors, which we really shouldn't be doing. So this value will be wrong.
Anyway. You want voltage. This means we have to compute the impedance at node Vo, to convert current into voltage. Bad news is this impedance is difficult to compute, since two collectors are facing each other. It will depend on individual transistor characteristics, ie their parapeter ro (due to Early effect).
What you are seeing is that the collector impedance of your transistors is high enough so that output voltage clips against one of the transistors.
This circuit should be used with a low input impedance stage after it, which takes a current as an input, and imposes a voltage to point Vo.