No the methods of *T would not be promoted. The specification doesn't explicitly allow it so it isn't allowed. However, there is a reason behind this.
At times you may call a *T method on T. However, there is an implicit reference taken. Methods of *T are not considered part of T's method set.
From the calls section of the Go specification:
If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m()
From the address operator section of the Go specification:
For an operand x of type T, the address operation &x generates a pointer of type *T to x. The operand must be addressable, that is, either a variable, pointer indirection, or slice indexing operation; or a field selector of an addressable struct operand; or an array indexing operation of an addressable array. As an exception to the addressability requirement, x may also be a (possibly parenthesized) composite literal.
If a S contains a *T, you don't even need to take its address so the methods can be called. If *S contains a T, you know T is addressable because T is a field selector of a pointer indirected struct. For S containing T, this cannot be guaranteed.
UPDATE: why does that code work?
Remember that *S contains the method set of *T. Also, as I quoted before:
If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m()
Put the two together and you have your answer. Counter is addressable and &counter contains the method set of *T. Therefore, counter.Inc() is shorthand for (&counter).Inc().