Start by calculating the power required just to heat the air. Power is energy per time. A useful unit of power is Watts, which is one Joule per second. Let's therefore figure out how many Joules it takes to heat one second's worth of air.
To do these calculations more easily, we convert to more sensible units. 75 °F to 475 °F is 400 °F rise, which is a 222 °C rise.
Next we need to find the mass of air that is being heated. You say the flow rate is 100 CFM, but didn't specify if this is before or after the heating. I'll pick before the heating. 100 cubic feet = 2832 liters. 2832 l/minute = 47.2 l/s.
We need to convert that to a mass flow rate. 75 °F is 297 °K. A mole of ideal gas has a volume of 22.4 l at 273 °K, so 24.4 l at your intake temperature. Air is basically N2 with a molecular weight of 28, and O2 with a molecular weight of 32. Since N2 dominates, let's say air averages to a molecular weight of 29. That means a mole of "air" has a mass of 29 grams. The 47.2 l we are trying to raise the temperature of therefore has a mass of 56 g.
So now we have reduced the problem to finding the energy it takes to heat 56 g of air by 222 °C. Heat capacity of "air" is a squishy value since air isn't a constant. By digging around on the net, it seems 1.03 J/(g °C) is about right for this case. (222 °C)(56 g)(1.03 J)/(g °C) = 12.8 kJ. That's the energy required each second, so the power is 12.8 kW. With the various approximations and slop factors, three digits implies too much accuracy, so 13 kW.
The above is only the power going into heating the air, and doesn't account for any losses. If you insulate the outside of the pipe and heater well, let's say you need 15 kW. At 120 V, that would require 125 A. That's rather unwieldy, so I'd go to plan B and use 240 V, which requires 63 A. That's still a lot. At 15 kW, you should at least look into using 3-phase power directly.