Why is std::is_aggregate<T> an aggregate?

However, aggregate classes must not have a base class by definition

This is no longer true. [dcl.init.aggr]/1 defines an aggregate as

An aggregate is an array or a class with

• no user-provided, explicit, or inherited constructors ([class.ctor]),

• no private or protected non-static data members (Clause [class.access]),

• no virtual functions, and

• no virtual, private, or protected base classes ([class.mi]).

[ Note: Aggregate initialization does not allow accessing protected and private base class' members or constructors.  — end note ]

There is no longer a condition that it does not have a base class like it did in C++14 and earlier. As long as it has a public, non virtual base class that is now allowed. This means that the type traits are now considered aggregates as long as the above holds true for them

Since C++17, classes with non-virtual, not private or protected bases are aggregates: https://en.cppreference.com/w/cpp/language/aggregate_initialization