Will equilibrium ever establish if one of the products starts to disassociate?
Let's make this an abstract problem with species A, B, C, and F:
$$\ce{A &lt;=&gt; B + F}\tag{1}$$
$$\ce{B &lt;=&gt; C + F}\tag{2}$$
We can also write down the sum of the two reactions, which will also be at equilibrium:
$$\ce{A &lt;=&gt; C + 2 F}\tag{3 = 1+2}$$
And let's use a for the concentration of A and so on.
Finding the solution using reactions 1 and 3
The concentration of F is equal to the equilibrium constant for reaction (1), so the concentrations of A and B have to be equal:
$$a = b$$
For every mole of B made, we also get a mole of F. For every mole of C made, we get two moles of F. We know the concentration of F is 2.5 M, so we get:
$$b + 2c = 2.5 M$$
Finally, we started with 4 M of A, which makes equimolar B and makes equimolar C, so the sum of all three has to be 4 M:
$$a + b + c = 4 M$$
Now we have three equations and three unknowns. We can combine the first and third equation to eliminate $a$:
$$ 2b + c = 4M$$
We can multiply the second equation in preparation for solving for $c$:
$$ 2b + 4c = 5 M$$
Subtracting the latter equation gives the solution for $c$:
$$ c = \frac{1}{3} M$$
$a$ and $b$ come out as 1.833 M.
Le Chatelier

Lets say the first reaction reached equilibrium and then $\ce{N_2O_3}$ started decomposing. This means more amount of $\ce{O_2}$ will be produced and the equilibrium for the first reaction will get disturbed

Without the second reaction, the first will attain equilibrium when $a$ is about 1.85 M. The second reaction makes F and uses up B, so the first reaction almost stays at equilibrium, and the concentration of A does not change much.
Will it ever stop?

Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $\ce{N_2O_3}$ will decompose further to form oxygen.
But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

If you look at this step-wise (figure out how much one reaction would go, disregarding the other), you are never done. This is similar to Zeno's paradox. However, the corrections get smaller and smaller, and your approximation gets better and better.
In general, reactions approach equilibrium, they don’t reach it. Nothing special about this system.