Making a template parameter a friend?

It is explicitly disallowed in the standard, even if some versions of VisualStudio do allow it.

C++ Standard 7.1.5.3 Elaborated type specifiers, paragraph 2

3.4.4 describes how name lookup proceeds for the identifier in an
elaborated-type-specifier. If the
identifier resolves to
a class-name or enum-name,
the elaborated-type-specifier introduces
it into the declaration the same
way a simple-type-specifier introduces
its type-name. If the identifier resolves
to a typedef-name or a
template type-parameter,
the elaborated-type-specifier is
ill-formed. [Note: this implies that,
within a class template with a
template type-parameter T, the
declaration friend class T; is
ill-formed. ]

I recognize the code above as a pattern to seal (disallow the extension of) a class. There is another solution, that does not really block the extension but that will flag unadvertidly extending from the class. As seen in ADOBE Source Library:

namespace adobe { namespace implementation {
template <class T>
class final
{
protected:
   final() {}
};
}}
#define ADOBE_FINAL( X ) private virtual adobe::implementation::final<T>

with the usage:

class Sealed : ADOBE_FINAL( Sealed )
{//...
};

While it allows extension if you really force it:

class SealBreaker : public Sealed, ADOBE_FINAL( Sealed )
{
public:
   SealBreaker() : adobe::implementation::final<Sealed>(), Sealed() {}
};

It will restrict users from mistakenly do it.

EDIT:

The upcoming C++11 standard does allow you to befriend a type argument with a slightly different syntax:

template <typename T>
class A {
   // friend class T; // still incorrect: elaborate type specifier
   friend T;          // correct: simple specifier, note lack of "class"
};