For a 3.3v LVTTL input pin can a 1k pulldown resistor value be used.

 

you should be Ok. A value of 2.2K was common for many applications, but it really depends on what the fan out capacity is of the device that is driving the signal. Are you sure your not looking for a pull up?

 

To find the maximum value pull-down resistor for a logic IC unused input, look at the datasheet for "input currrent" and based on this current calculate a resistor whose value is below the worst-case input "low" voltage for the device.

If the input current is 1uA and the "input low voltage" is 0.8V, then the maximum value resistor would be less than 800K = 0.8V / 1uA. However, it is advisable to stay well below the input low voltage and the typical 1K will obviously meet the spec. For a driven input signal, then the ability of the driver to source current must be taken into consideration for determining the lower end of this value.

For automated board testing of an assembly, it is/was preferred to have a pull-down (rather than direct ground) so that all unused inputs could be driven independently to check for proper operation of the IC.

 

It appears that LVTTL is actually made with CMOS technology, and as such has negligible input current, so you can pull down an input with almost any resistance value, from zero ohms(if not driven), to Megohms. Personally, I would probably stay below 100k.
If the pin is driven high and pulled down with a resistor, then the node capacitance and the resistance will dictate the fall time.

EDIT: I couldn't find any LVTTL products that were made with a bipolar process, but I could be wrong about them all being made in CMOS.