There are several object-oriented languages where one can define new root classes, including C++, PHP, and Objective-C, and they work fine, so this is definitely not a special thing.
There is a reason why Objective-C has a universal base class
As Sulthan mentioned, this is not true. There are multiple root classes in Objective-C, and you can define a new root class by simply not specifying a superclass. As Sulthan also mentioned, Cocoa itself has several root classes, NSObject, NSProxy, and Object (the root class of Protocol in ObjC 1.0).
The original Objective-C language was very flexible and someone could in theory come along and create his own root class and create his own framework that is completely different from Foundation, and uses methods completely different from retain, release, alloc, dealloc, etc., and could even implement a completely different way of memory management if he wanted. This flexibility is one of the things so amazing about the bare Objective-C language — it simply provides a thin layer, all the other things like how objects are created and destroyed, memory management, etc., can all be determined by the user frameworks sitting on top.
However, with Apple’s Objective-C 2.0 and modern runtime, more work needed to be done to make your own root class. And with the addition of ARC, in order to use your objects in ARC, you must implement Cocoa’s memory management methods like retain and release. Also, to use your objects in Cocoa collections, your class must also implement things like isEqual: and hash.
So in modern Cocoa/Cocoa Touch development, objects generally must at least implement a basic set of methods, which are the methods in the NSObject protocol. All the root classes in Cocoa (NSObject, NSProxy) implement the NSObject protocol.
So, what’s up with that? Are Swift classes with no defined
superclasses just NSObjects that pose as proper root classes under the
hood? Or is the default object-behaviour duplicated for each new
root-class? Or have they created another Swift-baseclass?
This is a good question, and you can find out by introspection with the Objective-C runtime. All objects in Swift are, in a sense, also Objective-C objects, in that they can be used with the Objective-C runtime just like objects from Objective-C. Some members of the class (the ones not marked @objc or dynamic) may not be visible to Objective-C, but otherwise all the introspection features of the Objective-C runtime work fully on objects of pure Swift classes. Classes defined in Swift look like any other class to the Objective-C runtime, except the name is mangled.
Using the Objective-C runtime, you can discover that for a class that is a root class in Swift, from the point of view of Objective-C, it actually has a superclass named SwiftObject. And this SwiftObject class implements the methods of the NSObject protocol like retain, release, isEqual:, respondsToSelector:, etc. (though it does not actually conform to the NSObject protocol). This is how you can use pure Swift objects with Cocoa APIs without problem.
From inside Swift itself, however, the compiler does not believe that a Swift root class implements these methods. So if you define a root class Foo, then if you try to call Foo().isKindOfClass(Foo.self), it will not compile it complaining that this method does not exist. But we can still use it with a trick — recall that the compiler will let us call any Objective-C method (which the compiler has heard of) on a variable of type AnyObject, and the method lookup produces an implicitly-unwrapped optional function that succeeds or fails at runtime. So what we can do is cast to AnyObject, make sure to import Foundation or ObjectiveC (so the declaration is visible to the compiler), we can then call it, and it will work at runtime:
(Foo() as AnyObject).isKindOfClass(Foo.self)
So basically, from the Objective-C point of view, a Swift class either has an existing Objective-C class as root class (if it inherited from an Objective-C class), or has SwiftObject as root class.