Linear regulator power consumption
The power lost in a linear regulator is equal to the voltage drop between input and output times the output current plus the input voltage times the current flowing to ground through the linear regulator.
\$P = (V_{in}-V_{out})I_{load} + V_{in}I_{gnd}\$
When there is no load, we have
\$P_{quiescent} = V_{in}I_{quiescent}\$
When there is full load, it is generally the case that
\$I_{load} &gt;&gt; I_{gnd}\$
and we can often safely ignore \$I_{gnd}\$
giving us
\$P_{loaded} \approx (V_{in}-V_{out})I_{load}\$
You mention in a comment that the highest \$V_{in}\$ will be 4.5V. In that case, the voltage drop from \$V_{in}\$ to your load will be 3.0V. Multiply that by \$I_{load}\$ and you will have approximately your maximum power consumption of your regulator.
If your input voltage drops to exactly 1.5V, even a low drop-out regulator will not deliver exactly 1.5V under a non-trivial load. In trying to maintain output voltage, \$I_{gnd}\$ may increase significantly above \$I_{quiescent}\$. You may want to choose a voltage detector whose tripping voltage is above your regulator IC's minimum suggested operating \$V_{in}\$ to avoid that problem.
According to the XC6125 Datasheet p.33, the dropout voltage at 1.5V output may be as high as 890mV. So, you may want your voltage detector to trip at 2.390V to avoid operating below the drop-out voltage range.