The voltage across an LED is (almost) constant for currents close to the nominal value. Assuming it's driven from a constant voltage source (eg. wall wart, battery), this means that the voltage across the series resistor will also be constant. Power dissipated in the resistor is V2/R. V is constant and R increases with a higher value resistor, so power dissipation will go down. In addition, power dissipated by the LED also goes down. This is a simplification that may not be accurate if you vary the current over a large range (say more than an order of magnitude), in that case you'd have to take the exponential IV curve of an LED into account, but the total power would still go down.