13.7.5 [temp.friend] bullet 1.3 says:

• if the name of the friend is a qualified-id and a matching specialization of a function template is found in the specified class or namespace, the friend declaration refers to that function template specialization, otherwise,

I'm not sure this says what it's supposed to say. For example:

    namespace N {
        template<class T> int f(T);
    }

    class A {
        friend int N::f(int);
        int m;
        A();
    };

    namespace N {
        template< class T > int f(T) {
            A a;            // ok for T=int?
            return a.m;     // ok for T=int?
        }
    }

    int m = N::f(42);       // ok?
    char c = N::f('a');     // Clearly ill-formed.

The key is that the wording talks about a “matching specialization,” which to me means that N::f<int> is befriended only if that specialization existed in N before the friend declaration. So it's ill-formed as written, but if we move the call to N::f<int> up to a point before the definition of A, it's well-formed.

That seems surprising, especially given that the first bullet does not require a pre-existing specialization. So I suggest replacing bullet 3 with something like:

• if the name of the friend is a qualified-id and a matching function template is found in the specified class or namespace, the friend declaration refers to the deduced specialization of that function template, otherwise,