Hallo Erik,

of course, if I would build such an extreme arrangement, I would measure capacitance of each used capacitor. And I would take only those ones, which do show smallest deviation from nominal value. But tolerances of today fabricated electrolytic capacitors are much smaller than in former times. Also, fabrication of big capacitors is much easier than that of very small ones, so that tolerance can be made quite smaller.
If you take a modern type manufactured by 'epcos', 'rifa', 'panasonic', 'bhc aerovox', etc. you will normally find a tolerance of +-20%. Sometimes they will also have -10/+30%.

In the 'transformer solution' suggested by me, 32 (3 x 32 arrangement) or 42 (4 x 42 arrangement) of same capacitors are connected in parallel. Statistics tells, that deviation of capacitance of this parallel connection reduces by the factor 1 / SQRT(n), which is 1 / 5.7 or 1 / 6.5. So, each block being connected to a winding of transformer shows only a deviation +-3.5% or +-3.1% from expected value, when +-20% tolerance parts are used. Even if -10/+30% parts are used, deviation from expected value will only be -1.8/+5.3% or -1.5/+4.6%.

Each block of 32 capacitors connected in parallel is connected to exactly the same voltage. So, this does not mean any problem. Only capacitance is a bit away from expected value.

Connecting such blocks in series CAN present a problem, as Michael Neary stated, if capacitor array is discharged totally by this 1200A current. But this is not planned to do. Nevertheless, if this cannot be exluded to happen, a protection circuit should disconnect the load from capacitor array, when certain undervoltage is detected, perhaps 960V, which represents 20% undervoltage.
Also anti-parallel protection diodes (of course the big stuff!) across each block could help to prevent reverse voltage at capacitors.

Assume 3 blocks are used, each containing 32 capacitors of 4700µF/450V. If the upper two blocks show +5% tolerance, and the lowest block -5%, then maximum reverse voltage can be estimated by this:

Upper two blocks' capacitance is 32 x 4700µF x 1.05 = 157920µF. Lowest block's capacitance is 32 x 4700µF x 0.95 = 142880µF. Total capacitance is 1/(1/157920µF + 1/157920µF +1/142880µF) = 50856µF.
Total discharge with constant current 1200A lasts dt = dU x C / I = 50.856msec.
Discharge of 157920µF capacitance by 1200A for 50.856msec yields dU = I x dt / C = 386.44V. Discharge of 142880µF capacitance yields dU = 427.12V.
But because each block was once charged to 400V, upper blocks still contain 400V - 386.44V = 13.56V, while lowest block has negative voltage of 27.12V.
This case must be prevented, of course!!

You stated, that new capacitors keep on being formed, so that capacitance will rise. This mechanism, which is better called 'regeneration', is represented by 'initial leakage current', which has totally stabilized after about 30min. Already after 5min initial leakage current has become very low and is only about twice the 'operational leakage current'.
From rather low value of initial leakage current and from rather short time of this duration I don't think, that this results in any relevant increase of capacitance.

Nevertheless, there's also a capacitance drift during service life, of course, which is specified to be within +-10% at end of service life. One might think, that 1200A will cause a much heavier shift of capacitance. But keep in mind, that this current distributes over 32 or 42 capacitors, which results in less than 38A for a duration of only some milliseconds. If you have a look at PEH 200 series of 'rifa' you will find, that rated current at 40°C and 10kHz is about 43.8Aeff permanently!! So, spike current of 38A lasting for only some milliseconds is not at all a special stress, just standard task, what the capacitor is built for.

Statisitics tells, that influence of capacitance drift will also be reduced by 1/SQRT(n), so that +-10% long term drift will only present an additional shift of less than +-1.8% per block, which is quite acceptable.

Kai