It seems this would only be a risk if the very last hashing function were compromised. Even then you would need to know how to generate the correct input to the last hashing function using the other hashing functions. It seems inherently more secure to me.
if hash1(x) has collisions, then so does hash2(hash1(x)) and hash4(hash3(... and so on, until we reach the full hashing stack, which also has collisions. Similarly, if hash2, hash3, etc, or hash11 have collisions, then so does X11(x). Simply put, if there's a collision attack for any hash#(x), then the same attack applies to X11(x).
This is the same baseless declaration with more words. You still failed to make the correlation or demonstrate a causation.
Why would this result in what is essentially a cascade failure? Why would an attack on Math A result in a failure of all other Math B to Math K to be Math anymore?
And on top of it, it's still an if...
I think you're fundamentally wrong. Collision on hash A does not break Hash B.