So...you have 10 constant-current circuits in series, so that each of them can contain 1 diode. You might have some issues with that because they must all have exactly the same current simply because they're all in series, but they'll all be set slightly differently because no two components are ever equal.
Therefore, one circuit (which one is completely random) will do all the work of regulating the entire chain while the rest become much closer to short-circuit because they're not quite getting the current that they're set for. Now that one circuit is dropping nearly ALL of the voltage left over between the LED string and the power supply, it's going to get stressed possibly beyond its design.
It would be a much better idea to take your single circuit and just give it all 10 diodes in series. In total: 2 transistors, 2 resistors, and 10 diodes. R2 (current-sense) and the lower transistor can stay exactly as they are, R1 (transistor bias) should be increased by a yet-unknown amount, and the upper transistor should be able to handle the entire residual voltage at the intended current.
Now then, about ripple:
Because you have a regulator - specifically a current regulator, as opposed to the more common voltage regulator, but it's still a regulator - you really don't need to care that much about ripple. You have a minimum voltage to do what you want (dropout voltage), you have a maximum voltage from the rectifier, and in this case you have a well-known operating current.
Choose a cap or build a cap pack that can handle ripple current equal to your running current, and will still be above the dropout voltage just before the next rectifier pulse while supplying that current.
Once you figure that out, double the capacitance or more to account for component tolerance and product life expectancy (electros lose capacitance over time), order it, and test it.