As I see it, access to field f is not granted.

8.5.3 says "Access to a member of a type is permitted only if all three
of the following conditions are met:
1. The type is visible.
2. The member is accessible.
3. All relevant security demands (see clause 8.5.3.3) have been granted."

Back to your example:

public class Outer {
    internal class Inner {
        public Foo f;
    }
}

Let's check the visibility.
8.5.3.1: "Nested within another type. In this case, the type itself has
the visibility of the type inside of
which it is nested (its enclosing type)"

So, you are right: Inner has the same visibility as Outer. f is public,
its type is visible, ignoring security demands, it would be accessible
according to 8.5.3.

However, check the accessibility of Inner. The important thing is that
Inner is treated like a member here, this is mentioned in 8.5.3.4:
"A nested type has the same visibility as the
enclosing type and has an accessibility as would any other member of the
enclosing type. This accessibility
determines which other types may make references to the nested type."

So, although access to f seems to be granted, its containing type cannot
be referenced. When you cannot reference the type, you cannot reference
the field.

That's how I see it.
Fabian