No, not constant current but you can use it as a voltage dropper in the same way as the capacitor and it will then pass DC. The problem is the size of the inductor which could be physically large and it could be quite heavy. The DC current will only be limited by the resistance of the wire which will almost certainly be too low and you have another problem - when you switch off, the magnetic field in the inductor will collapse and could produce a VERY high voltage, possibly several KV which could damage the LED and surrounding circuitry.
You can work out the values needed with these formaulae:
For a capacitor, the effective resistance is Xc = 1/(2 * pi * f *C)
For an inductor, the effective resistance is XL = 2 * pi * f * L
In both cases 'f' in in Hz, C is in Farads and L is in Henries. You can work out what value for Xc or XL you need using Ohms law, X = V/I where V is the voltage you want to drop (input - LED voltage) and I is the LED current.
Brian.