Is `→` a type operator?
Yes, these can all be viewed as operators at the type level, but they're not all completely analogous. Most of these are type constructors in that they're formation operators for types, though type-level application is simply a type operator.

$\to$ may be viewed as an ordinary type constructor. In Pierce's notation, this is denoted $\lambda X . \lambda Y . X \to Y$. You can just as well think of it as binary algebraic structure on the set of types $(-) \to (-) : \star, \star \to \star$, where $\star$ is the kind of types. (If you have a higher kind above this, it could live there, but this depends on precisely what your calculus is.) The product of types, $\times$, may be viewed similarly. Type-level application is also a type operator.

$\forall X$, $\exists X$ and $\lambda X$ are also type constructors, but they're a little different from the previous two in that they're binding operators: just like the $\lambda$ term operator, they bind a variable in the context (though in this case, a type variable). This means these type constructors are no longer algebraic operators à la universal algebra like the simple type constructors $\to$ and $\times$.