I found this circuit, should solve your problem.
Good luck. LED- one-shot circuit (time adjustable)
Only have to add a resitor in serie,to reduce the voltage over your LED.
(between output (pin 3),LED and ground.)
That would only work if you had a double pole momentary wouldn't it? It seemed like he is trying to have the power button/ psu power activate the circuit. The mobo power on signal needs to be on a different ciruit than the timer circuit. Anyone know why the switch on that timer circuit must be momentary? Or could it just be tied to ground and turn on when the power comes.
You can use a power on reset circuit, if you want it to happen at power-up. The site explains the theory of operation well. The node between the resistor and the capacitor will rise in voltage as indicated at the site.
Having that node tied to the input of a TTL inverter (like a 74LS04, for example), will give you a step to on/off as the voltage crosses the gate's threshold.
You can adjust the timing by choosing values for the resistor and capacitor using the RC constant.
The button that turns on your computer simply grounds an input on the PSU. If you want to use the 555-based circuit Lite_User recommended, there's no reason you can't use the same button to trigger your 555.