Defining weight as the "gravitational force acting on an object", and disregarding the minimal impact that gravity has on objects considered to be 'gravitationally unbound', do all objects in space have a weight?
For example, the Sun exerts a gravitational pull on itself. Does this mean that it has a weight? If so, how much does it weigh?
Mass is the amount of stuff in an object. An object always has mass.
Weight is the gravitational force a nearby planet exerts on an object. The object has weight only if there is a planet nearby. (Not just a planet. A star works. Or any other object with mass. Two brick attract each other. But for ordinary sized objects, gravity is so weak you don't notice it.)
Usually one speaks of weight for an object sitting on the surface of the Earth. The object is at rest because the floor exerts an upward force that balances the weight. The upward force on your feet give you the feeling of weight. You need muscle to stand upright.
If an object is falling, there is only gravity. Gravity acts on every part of your body. You can relax. You don't need muscle to stay straight.
This is why free fall is called weightless. It is confusing because Weight, the force of gravity, is still acting on you.
In free fall, usually one speaks of the acceleration from gravity, not the force.
There are two ways you can think about the Sun.
One part of the Sun attracts other parts. You think this way when thinking about what holds the Sun together. It helps you figure out things like the pressure at the center of the Sun. So each part of the Sun has a weight.
The Sun as a whole attracts the Earth. Here you think of the Sun as a single object with a mass that is the total of its parts.
You could figure out the force from each part of the Sun and add them up. You would get the same answer for the total force on the earth.
The weight of the Sun from its own gravity doesn't really work. Weight is a force. Force is a vector. If you add up the force on all the different parts, you get $0$. You know this from Newton's laws. Each pair of parts exert equal and opposite forces on each other. The total for each pair is $0$.
This question is really mixed up. Is it about "weight" or is it about gravitational self interaction?
If it is about weight, then it is meaningless, but the fact that it is so is an important part of understanding gravitation, so it is a good question.
For an object, let's take a baseball. ($m=145\,{\rm g}$). It has a weight here on Earth, on average:
$$ \bar w = mg= (145\,{\rm g})(9.80665\,{\rm \frac m{s^2}}) = 1.42196425\,{\rm N}$$
Meanwhile, if Santa Claus is prepping his sled so Lil' Timmy can get that brand new Big League Baseball for Christmas, that ball weighs in at:
$$ w_{\rm North\ Pole} = 1.42564\,{\rm N} $$
Up at the ISS, if Butch can't take it anymore and heads out to play catch with his stranded partner, Sujnita, they are weightless and the weight of the ball is 0.
Back in the ISS, there is about 1 micro-gee from drag, so the ball weighs:
$$ w_{ISS} \approx 1.42196425\,\mu{\rm N}$$
Meanwhile, in Earth Centered Inertial coordinates, the ball on the ISS weighs almost its full 1.42 newton, there is a small reduction for the increased radius from Earth. It's just on a ballistic arc. Whether there is Coriolis or Centrifugal force also on the ball--that depends on coordinates.
The point is: weight is entirely frame dependent, and whether that weight is from Newton's gravity, centrifugal force, or Coriolis force is also frame dependent.
Regarding the Sun pulling on itself--why not just use the Earth? If there is a 100 pound rock, that is part of the Earth, and it weights 100 pounds, at the surface. So the Earth is made of pieces of Earth that have weight.
Edit: I didn't get to finish the self-gravitation part, so here it is: you need to do an integral over the volume because ofc, gravity changes with radius from the center. For a uniform density, you get a nice result (compare to charge, for example), but the Earth isn't uniform, and the aforementioned 100 pound rock increase weight down in the mines (and beyond), for a surprising distance:
