Is it decidable whether a Turing machine will ever leave the start state on any input?
I would say decidable. Here is the idea.

Firstly, let your alphabet be $\Sigma$, including the empty symbol. I claim that unless you have in the transition function $(s_0, x) \to (s_i, \dots)$ for every $x \in \Sigma$ and $ i \neq 0$ your machine is not the desired one. Why this is true?
Suppose there is such $x$ for which TM doesn't change the initial state. Fill the tape with all these $x$'s. Done.

Here I had the assumption that we can have any initial tape as we want. If you consider only finite non empty input the problem becomes much harder.