When you're unsure about how Python treats an expression, then the ast¹ module is your friend:

    >>> ast.dump(ast.parse('n != x &~ n'), False)
    Module([Expr(Compare(Name('n', Load()), [NotEq()], [BinOp(Name('x', Load()), BitAnd(), UnaryOp(Invert(), Name('n', Load())))]))])

It can make it easier to reason through the curiosities, such as the precedence in the given example.

Note: Suggesting `ast` as a solution surely proves continuational's point ;)

1. https://docs.python.org/3/library/ast.html