Minimum specs before a flyback diode is needed for a relay?
When controlling inductive loads with semiconductor switches (transistors), the most dangerous for the devices is "flyback voltage" that is generated by an inductor when the switch gets turned OFF.

In theory the back-EMF voltage is determined by self-inductance and the rate of current change, as presented in layman terms:



VL = Ldi/dt

Therefore, the practical level of voltage spike depends on how fast your switch is. If you would use a mechanical on-off switch, when the switch impedance changes from milli-ohm (switch closed) to giga-ohm (switch open) in a fairly short time, the di/dt is infinitely high and the theoretical voltage spike is infinitely high. That's why electro-mechanical switches always "arching". And holding the inductor's wires might give you a sensible shock when disconnecting them from a battery. Therefore this flyback voltage can damage any silicon device. That's why the clamping diode is always a necessity. For engineering-level description, look at this Infinion appnote.

Technically you can reduce the di/dt using a very slow switch (with high breakdown voltage), and avoid the flyback diode, but slowing down the switch will likely cost much more than a diode. 

ADDITION: Quick LTSpice simulation of the OP circuit produces a spike across the transistor with amplitude of 11 kV (!!!). This is under the typical GPIO edge rate of 10ns, and the relay parameters were assumed as 25 Ohm coil, 90 mH inductance.

If a N-FET 2N7002 used, the spike goes up to 7 kV. Slowing the switch time by using 100k (!) gate resistor R1 produces the switch time of about 100 ns and the spike amplitude of about 200V, which is still too high for a 60-V transistor.

However, using a 1N4148 diode across the relay coil eliminates all spikes to 4-V level.