Power at the differential output
For a resistive load, if the voltage across the load increases by a factor of 2 and the load impedance stays the same, then the power into the load should increase by a factor of 4, or about 6dB.


$$ P_1 = \frac {E^2}{R} = \frac {1V^2}{1\Omega} = \text { 1 watt}  $$
$$ P_2 = \frac {E^2}{R} = \frac {2V^2}{1\Omega} = \text { 4 watts}  $$
$$ dBw = 10log{_{10}} \frac{P_2}{P_1} \approx 6dB $$


Another way of looking at it is that if the voltage across a load doubles, then the current through the load must also double and, since power is the product of current and voltage, the power into the load must quadruple.