How to calculate the conduction losses diodes in each stage
Assuming you are talking about a halfwave Cockcroft-Walton voltage mulitiplier, you can approximate the output voltage from this equation:
$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{2 \pi fC} (4n^3 + 3n^2 - n) - 2nV_{f}$$
What that does is to compute your output voltage from several parameters:

\$E_{pk}\$ the peak voltage of the AC input (not the peak to peak voltage, but the simple peak voltage.)
\$I_{load}\$ - the load current.
\$f\$ - the frequency of the AC input
\$C\$ - the capacitance of your individual capacitors (in farads.)
\$n\$ - the number of stages in your multiplier.
\$V_{f}\$ - the forward voltage of your diodes.

You can estimate the impedance from an \$n\$ stage multiplier with this equation:
$$ Z = \frac {4n^3 + 3n^2 - n}{2 \pi fC}  $$
I've posted the above equations before, and I have some background on them on my blog.
You can't really calculate a &quot;power loss&quot; for each stage.

The capacitors don't dissipate (much) power - it's reactance and not resistance.
The diodes will dissipate some power, but again not much.  You normally only draw milliamperes of current from a multiplier.  The product of the current and the forward voltage gives a dissipated power of milliwatts in the diodes.

The biggest problem with a voltage multiplier is the impedance.  As you can see, it goes up quite rapidly with the number of stages.  You can use larger capacitors and/or higher frequencies to make the impedance lower, but it still goes up rapidly.

If all you really need is the conduction losses in the diodes, then that is the product of the load current and the forward voltage of each diode.
$$ P = I_{Load} \times V_{forward} \times 2 \times n$$
That will tell you the power dissipated in all the diodes for an n-stage half wave Cockroft-Walton voltage multiplier.

Something you should not do is to short circuit a voltage multiplier.  At high current levels, the output voltage drops to such an extent that you have just a bunch of diodes in series between the source and the output.  If the source impedance is low enough, it will destroy the diodes.