If each LED has a 100 ohm resistor and is fed via 5 volts then the individual LED current is probably around 35mA. (Based on 1.5 volts LED forward voltage leaving 3.5 volts to be dropped across the 100 ohm resistor) So 28 LEDs will demand say 980mA.
To reduce the LED current using a single pot' (wired as a 2-terminal rheostat) you are anticipating that the individual currents will balance out (which they may or they may not - probably not)
Assuming they do and you want to reduce individual current to say 5 mA then total load current is 140mA. (28 x 5)
Individual LED series resistance to produce 5mA would be around 700 ohms. (3.5 / 0.005)
So your pot' will have to simulate 28 times 600 ohms in parallel, in other words 22ohms (600 = 700 - 100) (22 = 600 / 28)
So, you need a 22 ohm pot, rated for 140mA ie 0.43 watts. Let us say 0.5 watts
However, the wiper of A 22 ohm 0.5 watt pot is only rated to carry 140mA, and you want it to carry the maximum current of 980mA so in fact the total pot rating must be 3.5 x 0.980 = 3.5 watts (as you originally suggested)
However, note that the pot value for 28 LEDs is only 22 ohms. If you use more or less LEDs you need a different pot to achieve the same level of dimming.
So, your best bet is to use a 10k pot to produce a 0-5 volts into an arduino (or similar) analogue port, then use a PWM output to drive a FET rated at say 5 amps to drive as many of your LEDS as you wish. Irrespective of whether you are driving 1 or all 75, this will give you fully balanced brightness control from a single pot.