Easier ftl::Flags construction

As demonstrated by the existing test, while it is possible to construct
a ftl::Flags<E> from multiple values, it requires an extra
`using namespace ftl::flag_operators` or alternatively a
`using ftl::flag_operators::operator|` to implicitly construct a
`ftl::Flags<E>` type.

But there is an easier way -- allow implicit construction from a
`std::initializer_list<E>`.

This means as an alternative to:

    using namespace ftl::flag_operators;
    ftl::Flags<E> x = E::A | E::B;

... one can instead write:

    ftl::Flags<E> x = {E::A, E::B};

... and achieve the same initial value.

This change adds the new constructor overload.

Assignment from an initializer list automatically works, without having
to define a explicit `operator=(std::initializer_list<E>)`. Instead the
copy constuctor is used.

As a useful side effect, you can now also clear the flags by assigning
an empty initializer list:

    ftl::Flags<E> x;
    x = {};  // Clears x to zero

Bug: 185536303
Flag: EXEMPT New library code
Test: atest ftl_test
Change-Id: I1de7857c93ebef4fc5e6157aac9cf162b49943e1