The number of turns is irrelevant.
All changing the number of turns does, assuming you also change the wire diameter so that the resulting winding fills the space available, is to change the impedance of the winding, the voltage/current ratio, not the efficiency.
Consider the space filled with two identical windings, running at the same current, voltage, field, dissipation etc. If you connect them in series, it runs at 2V and I, you have twice as many turns as each coil. If you connect them in parallel, it runs at V and 2I, you have double the copper area but the same number of turns. All of the important parameters, field, weight, torque, power lost, cost of the copper, are unaltered, even as you've doubled or halved the number of turns.
That's all to first order of course. There are many small second order things going on. The wires to the motor can be thinner if the motor impedance is higher. Thin wire tends to have a greater part of its cross sectional area occupied by insulation than thick wire, so it's not quite as efficient as medium weight wire. Very thick wire also doesn't pack as well due to its stiffness. A very high voltage motor may well dedicate more space to insulation than a lower voltage motor. There may be other considerations for voltage, for instance availability of 12v batteries, or staying below 40v so you can use 'low voltage' insulation rules.
That's enough about number of turns not affecting efficiency, so what does?
The main loss is \$I^2R\$ loss in the copper. Unfortunately there's little that can be done about the resistivity of copper. Silver is too expensive (and barely any better), cryogenics is very complicated, though superconductor is worth doing if the machine is really, really big (>> 100MW).
High field magnets produce more volts per turn, and more torque per current than low field magnets, so you should use the highest field you can afford. More volts per turn means a higher ratio of 'useful' volts (coupled to the speed of the motor) to 'useless' \$IR\$ drop volts that only generate heat, so better efficiency. A small airgap helps produce a higher field with the same magnets, but requires better machining tolerance and produces more air viscosity losses (aka windage), so it can't be reduced too much.
A high speed motor produces more power than a low speed one at the same output torque, which is another way of saying generates a higher EMF than a slow one. Again, a higher ratio of useful volts to useless volts. However, high speed requires strong materials, and careful balancing, and produces losses that increase rapidly with speed, so it can't be increased too much. The higher rotor frequency means more hysteresis loss in the magnetic materials, and the windage losses increase as the speed cubed, so will rapidly come to dominate at very high speeds.
Because of the way some power and loss terms scale, a big motor, other things being equal, will have lower losses than a small motor. But let's say you want four driven wheels. Is it better to use a small motor per wheel, or a big motor and a mechanical drive train? Almost certainly the former, for lower weight and complexity, even if it's slightly less efficient.
Motor/generator design is a compromise between efficiency and, well, cost really, once you have a motor powerful enough for your specifications. Unfortunately, even the simple things like cost, weight and volume are a multidimensional space you can't simply optimise over. When you add things you can't easily put numbers on, like time to design, maintainability, reliability, the rest of the vehicle efficiency, convenience, use of strategic materials (in strong magnets), there is no one size fits all, no one equation describes all.