NPN Transistor with Diode logic
Try thinking about the situation like this.
The transistor has a load connected to the collector, it is not aware of the nature of the load but it will try and maintain the collector current in accordance with the base current as you have described i.e. Ic = hFe * Ib. The gain might well be 100x for small base currents.
As the base current is increased eventually you'll reach a point where the collector current is not increasing proportionally. The transistor is going into saturation. The gain may be only 5x (for example) at this point and Vce will go down to a small voltage say 0.2V
In your case the circuit will be dropping 8.8V across the load (diode and resistor). With a diode forward voltage of 1.2V there will be 7.6V across the 325ohm resistor or about 23mA through the load. Driving the base with a higher current will not change this as the transistor is saturated.
Regarding the transistor's behaviour i.e. saturated as being independent of the load we could get a better estimate of the collector current by either graphically plotting the diode characteristic curve against the resistor volt drop or using the diode equation to solve iteratively. Either way you are correct there will be about 23mA through the LED. You can verify with the LED's datasheet to check it can handle this current and check the forward voltage.