In this circuit, the large capacitor will start at ~4.7V (5V minus V-D1, here I assume 0.3V but be careful if you have a generic 0.7V diode, and also beware of the voltage divider effect of R3) and decay from there, so it will be a game between the 4.7V sagging down and when the RPI browns out. If this was a production design, then you are spending way too much on the huge capacitor. Also, if the capacitor is fully discharged, then be aware of the inrush current that will occur when it is first powered up. With such a huge capacitor, D1 and R3 will see a ton of amperage for a bit of time.
So if you make D1 and R3 big enough (in terms of wattage), and the brown-out level for the RPI is well below 4.7V, and R3 isn't too high of impedance, then it will work. But I wouldn't do this in a production product.
It would be more efficient to connect your 12->5V converter to the battery (use a fuse). That way, the RPI isn't racing against time. You could put in a hold-up circuit so that when the RPI detects the car is off, it can do its tasks and then flip an output that turns off its own power, thereby conserving your car battery. No worries about brown-out behavior, minimal inrush current, no rush against time for the RPI, minimal drain on battery, and much cheaper in component cost.