How can I make tuples an instance of this class in Haskell?

(x,y) is already a concrete tuple type, containing two concrete (albeit unknown) types x and y. A functor, or bifunctor, meanwhile is supposed to be parametric, i.e. in case of the tuple instance you'd want the contained types to be left open as parameters, which are then filled in with various different concrete types when the methods are used.

I.e., you basically want a type-level lambda

instance Bifunctor (\x y -> (x, y)) where

Well, Haskell doesn't have type-level lambdas, but it does have partial application at the type level – in this case not even partial, you don't want to apply the tuple constructor to any types at all but simply leave them open. That is written thus:

instance Bifunctor (,) where

If you wanted to apply it to only one argument, you write

instance Functor ((,) a) where

which I find easier to understand if parsed as Functor (a,) – but that's not actually legal in Haskell.

Good question.

The class applies to the functor type itself, and in your case the functor type is (,). To get the feeling about it, notice the difference here.

:t (,)
(,) :: a -> b -> (a, b)

:t (True,False)
(True,False) :: (Bool, Bool)

It would probably have been more intuitive if you had used a Pair type like :

data Pair a b = Pair a b

Because reading the class definition would have made more obvious the type application of 'p'.

Just like Haskell uses types for values, as illustrated above, it uses types for types (also for compile-time logic), which are named Kinds.

:k Pair
Pair :: * -> * -> *

:k (,)
(,) :: * -> * -> *

:k (Bool,Bool)
(Bool,Bool) :: *

:k Bifunctor 
Bifunctor :: (* -> * -> *) -> Constraint

This last line illustrates that Bifunctor class is designed for types of kind (* -> * -> *), not kind (*) of (a,b), hence the error message you had from GHC.

Your definition was almost right, here is the correct one :

instance Bifunctor (,) where
  bimap func func' (x, y) = (func x, func' y)
  first func (x, y) = (func x, y)
  second func (x, y) = (x, func y)

EDIT : illustration of kinds as suggested by @leftroundabout