Is it possible that the halting problem is solvable for all input except the machine's code?
Recall the standard proof of the undecidability of the halting problem. Suppose that some machine&nbsp;$H$ decides the halting problem and let $Q$ be the machine that, on input&nbsp;$\langle M\rangle$ uses&nbsp;$H$ to determine if $M(\langle M\rangle)$ halts and, if so, $Q$&nbsp;loops; otherwise, $Q$&nbsp;halts. Now, $Q(\langle Q\rangle)$ halts if, and only if, it doesn't halt.


  Is it possible that the halting problem is decidable for any TM on any input so long as the input is not the TM itself?


No. If you change the definition of the halting problem in this way, the proof still works. We don't care what happens when&nbsp;$H$ receives&nbsp;$\langle H\rangle$ as input because the contradiction comes after we give input $\langle Q\rangle, \langle Q\rangle$ to&nbsp;$H$.

Second, if you modified&nbsp;$H$ to detect that input, we could get the same contradiction by using any other machine&nbsp;$Q'$ that is equivalent to&nbsp;$Q$ in the sense that, for any input&nbsp;$w$, $Q'(w)$&nbsp;halts if, and only if, $Q(w)$&nbsp;halts. There are infinitely many of these and $H$&nbsp;can't detect all of them because it's undecidable whether two Turing machines are equivalent.