Electronic – Transistor question and is the circuit safe for Raspberry-Pi

An alternative strategy would be to power the Raspberry Pi continuously and use the ignition line to initiate the power down sequence. I've done that in the past but with systems where the exact solution wouldn't apply to a Pi but in general:

Use a DC-DC converter for the best efficiency, there are many examples around but the following is one example of something that would be convenient to use and it can supply 1A at 5V from a 6.5V to 32V input:

http://www.digikey.com/product-detail/en/V7805-1000/102-1715-ND/1828608

A car supply can be quite harsh, so you might want to use a 30V TVS diode across the input to protect against spikes with a chunky Schottky diode with the anode at ground and the cathode at the 12V input to protect against negative voltages along with either a normal fuse or a PTC resettable fuse in series with the connection between the car's power and your system. Otherwise you may be able to 'hack' a car to USB charger that should already have all that in place.

I'm not sure what a Raspberry Pi draws in normal idle mode, but presumably well under 500mA which is the maximum USB can supply and more likely 100mA. Say it's using 100mA at 5V that will be under 50mA at 12V using that circuit, a car battery is normally in the order of 50Ah so that would be around 20 days to drain the battery to 50%. If the car is in regular use there's probably no need to go any further, and you may just be able to leave it running and just turn off any peripherals you're not using.

Otherwise for detecting the ignition change either way and both informing the Pi it needs to shut down followed by removing power a minute later the most practical way is probably to use an external microcontroller that drives a FET. It could be done with discrete logic but you also need to make sure power is re-applied when the ignition goes high, so it's not an entirely trivial excercise but parts costs will be lower than using a large cap.

The LED has a voltage drop of 2v and a runs well on about 10-15ma, but how do I handle the capacitor in parallel?

The capacitor won't have any effect on the LED current after the circuit has been "on" for a few microseconds, so you don't need to worry about it when calculating the resistor value.

do I need a resistor between the optocoupler and B as the video would suggest?

You don't need a resistor.

You don't need to limit current through the optocoupler --- it won't draw any more current than what's stimulated as photocurrent by its LED (driven by the 15 V AC)

You aren't using the signal at the optocoupler's 'C' pin as a logic signal, so you don't need it to pull to a low voltage when the opto is active.

A current path into the 'B' pin could speed up the operation of the phototransistor, but for this kind of application, you don't need nanosecond response times.

It looks like there's a pathway, when the optocoupler is closed and the transistor is allowing current from E to C, that directly connects the 3.3v source to the GPIO pin without a resistor on the route (via the transistor) - is that a problem? Wouldn't it cause too much current to the GPIO pin?

Assuming you have the GPIO configured as an input (and this should be the power-up default), it will have a high input impedance, and won't allow more than a few microamps to flow into it.

A few k-ohm resistor would likely not change the behavior any and would protect you from overcurrent in the event you accidentally program the GPIO as an output and the driven value conflicts with the value generated by this doorbell circuit.