Is MALL + unrestricted recursive types Turing-complete?
If additive commutations are omited in MALL, it's easy to show that the size of a proof decreases with every cut-elimination step. If additive commutations are allowed, the proof is not as easy, but it was provided in the original “Linear Logic” paper. It's called Small Normalization Theorem (Corollary 4.22, p71), which says that as long as the contraction–promotion rule is not involved (which is the case in MALL) normalization holds. The argument does not rely on formulas themselves, they could be infinite (e.g. recursively defined).

It means that it's not possible to encode a promotion for type $\mu\alpha.\;I \;\&amp;\; A \;\&amp;\; (\alpha \otimes \alpha)$ in MALL, since it would allow for fix point combinators. Some additional recursion construct would be needed for that.

Note: I believe that it's possible to use MALL together with a coinduction principle (introduction of $\mu$'s dual) to keep the system normalizing and obtain a promotion for this encoding of $!A$. Allowing recursive types in MALL + coinduction would then make it Turing complete. But as long as MALL is considered alone, allowing recursive types is not a big deal.