How to start solving this type of exercise: Determine if $L$ is in $RE\setminus coRE$ or $coRE\setminus RE$ or $R$ or not in $RE\cup coRE$?
The line of thought that I find the most fruitful is: try to come up with a (semi)algorithm for the problem. If you succeed, great. If you don’t, it’s probably impossible.

Let’s apply it to your problem: determining whether a Turing machine accepts only palindromes. The trivial way to solve this problem is to go over all non-palindromes, and check whether your Turing machine accepts any of them. This shows that your problem is co-r.e., since you can enumerate all Turing machines which accept some non-palindrome. 

There doesn’t seem to be a matching algorithm for enumerating Yes instances of your problem, so your problem is probably not r.e. Indeed, Rice’s theorem shows that your problem isn’t recursive, and so we can deduce that it is not r.e.

With a bit more work, you can show that your problem is $\Pi_1$-complete, that is, every problem in co-r.e. can be (effectively) reduced to it. To show this, it suffices to show a reduction from the non-halting problem to your problem. In this problem, given a machine $M$ and an input $x$, the goal is to determine whether $M$ does not halt on $x$. Given an instance $\langle M,x \rangle$ of this problem, construct a machine $M’$ which goes into an infinite loop (or rejects) unless the input is $10$ (an arbitrary non-palindrome), in which case it simulates $M$ on $x$, and then accepts (if $M$ ever halts). The new machine $M’$ accepts only palindromes if and only if $M$ doesn’t halt on $x$.