Level shifter using BJT schematic
is Q2 always off?

No it is always on. What Q2 does is that it tries to make a (small) current flow into it's collector. That current tries to pull down the voltage at Vout.

Q5, Q2 and R2 are sort of a current mirror but a bad one. The current through R3 isn't copied 1 to 1 as in a "proper" current mirror. Instead since R2 is present the current through Q2 will be much smaller than the current through R3 and Q5. Across R3 we get a fairly constant voltage of VEE - 0.7 V. Since R3 is 1 kohm the current through R3 will be a couple of mA. As said the current through Q2 will be smaller than that, like 100 uA or less (100uA is just my guess, it is too much work to make that a more accurate number, that does not matter anyway for explaining how the circuit works). That 100uA causes the Vbe of Q5 to be not 0.7 V but slightly smaller like 0.6 V (= 0.7 V - 100 mV, as 100 uA through R2 gives 100 mV).

That 100 uA pulls down Vout. Opposite is Q1 pulling the output up (through R1). If Vin is  high enough then Q1 can supply so much current that Vout will be pulled up, almost to the value of VCC.

When Vin has a very low voltage Q1 will be opened much less, supplying much less current so Q2 "wins" and Vout is pulled down.

Q5 is connected in diode mode, and its Vbe is fixed at 0.7V

Correct

Adding Q2 Vbe plus the voltage drop in R2 should then give 0.7V again

Indeed Vbe(Q2) + V(R2) = Vbe(Q5) = 0.7 V

What will happen there is that there will be a significantly smaller current flowing through Q2, R2 than there is flowing through Q5.

The above imply that Q2's Vbe is less than 0.7 OR the voltage dropo on R2 is zero.

Both are true, Vbe(Q2) will be slightly less than 0.7 V and there will be a small voltage drop (less than 100 mV) across R2.