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You're right that there is a problem here. You should look at Bonnafé's "Sur les caractères des groupes réductifs finis à centre non connexe" which is Astérisque 306. Specifically look at §9.D where this issue is dealt with. More generally, you've clearly got to a decent level with Deligne--Lusztig theory. I would highly recommend reading the early parts of Cédric's monograph, especially Chapter 3. It really clarifies some of the mess of Digne--Michel's text.
You can cite 17.12 and 17.13 of Lusztig's "Character Sheaves IV". From this it follows from the fact that the symmetric group has no Cuspidal two-sided cells.
The special unitary groups $\mathrm{SU}_n(q)$, with $n \geqslant 3$, always have a non-degenerate maximally split maximal torus. This is easy to see from a matrix representation. This is because all bar one of the elements are contained in the field $\mathbb{F}_{q^2}$, so one has enough wiggle room. However, when $q=2$ then one can check that the maximally split torus in ${}^3\mathrm{D}_4(2)$, ${}^2\mathrm{D}_n(2)$ and ${}^2\mathrm{E}_6(2)$ are all degenerate, even though they're non-trivial.
Yes, certainly! If $q=3$ and $F$ is split then for any root $\alpha$ if there exists a root $\beta$ with $\langle \alpha,\check{\beta}\rangle = -1$ then we have $\alpha(\check{\beta}(-1)) = (-1)^{\langle \alpha,\check{\beta}\rangle} = -1 \neq 1$ so the maximal torus is certainly non-degenerate. I didn't check the details but there should always be a root with this property.
@user61318 Well it was useful because I wouldn't have thought about the general case without it, so thanks for sharing it! For clarity, I should say that we are talking about the paper here arxiv.org/abs/1603.03914 which uses the techniques in the answer below to give a new proof of the results in the above mentioned paper of Regev.
Your formula is independent of $\mu$ so long as $m > 0$. I think there is a difference when $m=0$ due to the way that hooks are computed in the skew diagram versus a proper Young diagram. For instance a hook Young diagram of a partition of $n$ has a hook of length $n$ but a hook skew diagram whose shape is a partition of $n$ has no hook of length $n$. The Murnaghan–Nakayama formula indicates that these situations are therefore different.
You are right that the argument above is not correct. That is the same counterexample I came to in the end. This result is not so easy to prove correctly. A proper proof of this result using root data will appear in a forthcoming joint paper that I am working on. I was going to come back and update the answer once the preprint had appeared on the arXiv. In the meantime if you want to discuss this more you should just send me an email.
It means that for any field $K$ of characteristic 0 a simple $K\mathfrak{S}_n$-module is of the form $M \otimes_{\mathbb{Q}} K$ where $M$ is a simple $\mathbb{Q}\mathfrak{S}_n$-module. Moreover the map $M \mapsto M\otimes_{\mathbb{Q}} K$ is a bijection between the simple modules. From this one easily gets that for any field extension $K \subseteq L$, with $K$ still of characteristic 0, we have the map $M \mapsto M\otimes_K L$ is a bijection between simple $K\mathfrak{S}_n$ and $L\mathfrak{S}_n$ modules.
