"Centripetal force" does not exist, so there's no reason for it to obey Newton's 3rd law (or any other law, for that matter).
Ok, I will admit that this was not a good answer. I'm still confident it's correct, but I was too terse, and I didn't actually try to address OP's concern. (I am also pretty sure the downvoters agree with me, but I didn't state my point clearly.)
When people talk about the centripetal force on an object in circular motion$^1$, usually what they mean is the component of the net force on the object in the direction toward the center of the circle. Let's call this component $F_{\text{net},c}$. An object moving in a circle has to have a component $a_c$ of its acceleration directed toward the center of the circle, and we can work out from the kinematics of circular motion that $a_c = v^2/r$. Newton's 2nd law then forces us to conclude that $F_{\text{net},c} = ma_c$.
When we apply Newton's 2nd law, we add up all the forces acting on an object to get the net force,
\begin{equation}
\vec{\mathbf{F}}_{\text{net}} = \vec{\mathbf{F}}_1 + \vec{\mathbf{F}}_2 + \ldots \tag1
\end{equation}
The forces $\vec{\mathbf{F}}_1$, $\vec{\mathbf{F}}_2$, etc. may be all sorts of forces exerted by other, different objects on the object in question, like gravity, electric forces, magnetic forces, tension, normal/contact forces, friction, and whatever else.
When I say that centripetal force does not exist, what I mean is that we will never include it in the sum of forces on the right hand side of Eq. 1. The centripetal force is (a component of) the net force get when we add up all the actual forces acting on our particle, but it is not itself a force.
In some cases, there is only a single force $\vec{\mathbf{F}}_1$ on the right hand side of Eq. 1. For example, if you have a satellite orbiting a planet in circular motion, the only force on the satellite is the gravitational force from the planet, and as a shorthand you might want to refer to this gravitational force as the centripetal force.$^2$
In such cases, what we are calling "the centripetal force" does obey Newton's 3rd law$^3$, in the sense that, whatever body B is exerting the force $\vec{\mathbf{F}}_1$ on our object A, A will exert a force $-\vec{\mathbf{F}}_1$ back on B. Just like every other example of Newton's 3rd law, the two forces $\vec{\mathbf{F}}_{\text{B on A}} = \vec{\mathbf{F}}_1$ and $\vec{\mathbf{F}}_{\text{A on B}} = -\vec{\mathbf{F}}_1$ don't "cancel" because they act on different objects. Going back to our concrete example, A is our orbiting satellite, B is the planet it's orbiting, and the two bodies are exerting gravitational forces on each other with equal magnitude and opposite direction.
However, I find it highly problematic even in this simple case to say that the centripetal force obeys Newton's 3rd law. We now have two different meanings of centripetal force that we are liable to mix up. The first is $F_{\text{net},c}$, and the second is $\vec{\mathbf{F}}_1$, and it is not necessarily true that Newton's 3rd law applies to the first meaning. For example, even if the gravitational force from the planet is the only force on the satellite, maybe there is some other force acting on the planet that is keeping it in equilibrium. Then the net force on the planet is zero, and it is surely not the case that $(F_{\text{net},c})_{\text{planet}} = -(F_{\text{net},c})_{\text{satellite}}$.
In general, once we have to include more than one force on the right hand side of Eq. 1, talking about "the centripetal force" makes less sense. Consider the case of a marble rounding a vertical, circular loop. Both a gravitational force $\vec{\mathbf{F}}_g$ exerted by the Earth and a normal force $\vec{\mathbf{N}}$ exerted by the circular track act on the marble. Each of these forces individually obey Newton's 3rd law: the marble exerts a gravitational force back on the Earth and it also exerts a normal force back on the track.
Now, which of these forces is "the centripetal force"? Of course, it's neither of them, because what we really mean by centripetal force is the radially-inward component of the net force
\begin{align}
F_{\text{net},c} = N + F_{g,c}
\end{align}
and this is not any single force. It's a sum of components of two different forces exerted by different objects, and it is not the kind of entity that Newton's 3rd law has anything to say about.
I could go on, but hopefully I've cleared myself up and offered a better answer to the question. I would just like to offer a final remark. When teaching physics, we work hard to impress on our students that the words we use, like work, power, acceleration, and so on, have specific technical meanings. It seems ridiculous to me to then just ignore that and refer to a "centripetal force" which isn't actually a force at all. I am sure you can sympathize with me if you've ever had to field a question like "How come we don't have to draw the centripetal force on our free-body diagram?"
[1] As @GiorgioP points out in a comment, we can define centripetal force for more general motion by talking about the direction toward the center of the instantaneous osculating circle of our particle's path, but let's keep this simple and focus on purely circular motion.
[2] Of course, this is an idealization, and in real life there will be (at least) gravitational forces from every other mass in the universe that we need to account for.
[3] We have to be careful here because there are cases — for example, involving magnetic forces — in which Newton's 3rd law is violated, but the more general principle of momentum conservation is obeyed.
