As Bimpelrekkie explained, whether or not the voltage drop of the diodes is significant or not depends on what voltage you are working with.
Since I've been playing with these things for the last few days, I'll post some of the diagrams and measurements I've made.
I think the voltage traces explain fairly what happens to the output when the AC is not at its peak value.
Just FYI: The voltage I'm using (30 volts peak to peak) is kind of on the edge of making the forward voltage of the diodes I'm using not matter much. The forward voltage is about 3% of the available voltage, so noticeable but small enough to squint and and say "eh, close enough" most of the time.
For starters, the Cockcroft-Walton half wave multiplier you showed is built of a bunch Greinacher voltage doublers. The Greinacher is composed of a Villard doubler followed by a rectifier and a filter.
This is a Villard doubler:
It takes AC, and adds half the peak to peak voltage to the AC as DC. My transformer was putting out about 30 volts peak to peak AC. The Villard circuit added about 14V to that. I ended up with a peak voltage at about 30V. It isn't really DC. It is "fluctuating DC" with all of the voltage above 0.
The gray line across the middle of the image is 0 volts. As you can see, the AC was moved upwards and resulted in the "DC." The flat spots at the bottom of the DC come from the diodes. The diodes have a forward voltage of around 1 volt, so that's missing from the bottom of the curve.
Now here's the circuit for a Greinacher doubler:
This circuit is the basis of the Cockcroft-Walton multiplier. Look at your Cockcroft-Walton diagram, and kind of tilt your head to one side. I'm sure you'll recognize its "cells" as nothing more than a bunch of Greinacher doublers strung together.
This is what the output of a Greinacher doubler looks like:
It is a little bit lower than the Villard output, but it is much closer to DC. If I had used larger capacitors, then it would look exactly like DC.
Now, you string a bunch of Greinachers together, and you get a higher voltage. It would look a lot like that last picture, just with higher output voltage.
I'm getting about 24V out of a single Greinacher. If I put a bunch of them together to get a Cockcroft-Walton multiplier, then I get 24V for every stage I add to it.
1 stage 24 V
2 stages 48 V
3 stages 72 V
and so on.
You take the peak to peak voltage of your AC (30 V in my case,) subtract the diode forward voltage, and that's approximately the voltage you can expect in DC for a single stage. Multiply by the number of stages, and you have your (approximate) DC output voltage.
The next trick is figuring out the size of the capacitors to use. I used 100nF capacitors for my experiments (that was way too small when working with 50 hertz.) That makes my doubler have a very high resistance. A "load" of 1 megaohm noticeably drops the output voltage. I've been trying to figure out how to express the impedance of a Cockcroft-Walton multiplier as a function of the capacitance. I'm playing hooky from that right now. My plan was to build a couple of more multipliers with different capacitors and see how they respond to loading.