I'd say that what the ≈1.2 kW must be doing is field weakening of the motor. 1.2 kW is around 1% of the full power of the motor, so it's considered an acceptable loss. Though of course if you're costing downhill in neutral, it's a 100% loss: zero power out for 1.2 kW in. Even in drive "gear" at 70 mph, if it's say 15 kW to maintain that speed, that's 1.2/15 = 8% loss, and there would be other losses.
At low speeds, the permanent magnet is unopposed, giving full torque up to about 40 mph (I'm not sure of the exact speed, and it likely varies from model to model). From here on, I'll pretend that this threshold speed is fixed and is exactly 40 mph. I'll assume that the motor's constants are such that without field weakening it generates around 100 V per 10 mph, and I'll take the battery voltage as 400 V, making for easy math.
To drive at 20 mph, the motor controller bucks the voltage to half (simplifying a lot here), just over half to accelerate, just under half to regenerate. But without field weakening, anything past 40 mph causes strong regeneration until the speed drops below 40 again. To drive at 50 mph, the motor controller has to do some phase shifting that effectively opposes some of the permanent magnet's field. This reduces torque, but allows for higher speed. That's effectively an electronic gearbox, in fact a Continuously Variable Transmission (CVT): you can smoothly trade torque for speed, maintaining power. Power is of course the product of speed and torque. This is the motor's so-called "constant power region".
As an aside, field weakening doesn't have to cost any real power, just like a permanent magnet can provide a magnetic field with no power input. But field weakening requires current through windings that have resistance, so in practice it costs real power. If you had a superconducting motor, in theory that loss would be zero, at least in the motor itself, ignoring the power needed to maintain superconductivity. If I'm right, this answers an earlier question: where is that power going? It would be into the coolant loop ultimately dissipating in the small EV radiator.
So at 70 mph, you have to effectively buck around 4/7th of the magnet's field, otherwise you get that uncontrolled regen again. If I'm right, that's the reason that neutral is only available in ready mode; you need the battery, contactors closed, and motor controller doing its thing to keep the back EMF of the spinning motor from pushing into the battery via the free-wheel diodes in the motor controller. That's one of the drawbacks of a permanent magnet motor: yes, you get "free" magnetic field at low speeds, but you can't turn it off and you have to buck it at higher speeds. Tesla and a few other EV manufacturers generally use induction motors, which have no permanent magnets, but they still have to do something similar for field weakening, and they need to use power at low speeds to generate any field at all.
One way to test this is to check if the drain falls linearly to about 40 mph and then stays fairly flat to a standstill. If so, that confirms my theory. However if it's a straight line all the way down to zero speed, that's a count against my theory.
So in contrast to an ICE vehicle where neutral is a fairly passive state, an EV's neutral is actually quite active. I'm guessing that if the main battery is really flat, then the car may refuse to go into neutral, and it would have to be towed with the drive wheels off the ground. In four wheel drive EVs, that may imply needing a tilt tray. It should be pushable up to 40 mph, as the back EMF would be below the battery voltage. Towing much above 40 mph with the car "off" could damage the motor controller, since the un-bucked back EMF would be more than the motor controller is designed to operate at (around maximum battery voltage). In this case, with the car off, the motor won't regenerate, but the voltage could be approaching 700 V, and the motor controller might be using 650 V IGBTs (electronic switching devices), and the capacitors might only be rated at say 450 V or 500 V.