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Let

$a_{n+1}=3a_{n}^2(1-a_{n})$

Where $a_0$ is any complex number

For $f(x)=3x^2(1-x)$ I found fixed point $0,\frac{1}{2}(1\pm \frac{i}{\sqrt3})$

Setting $a_0$ equal to one fixed point we get $a_n$ constant for all $n$.

Is it possible to find a formula for $a_n$ ?

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There is a general technique for this situation. Let $\, f(x) := 3x^2(1-x).\,$ Now at the fixed point $\,x=0\,$ we have $\, f(x) = 3x^2 + O(x^3).\,$ We want a function $\,g(x)\,$ such that $\, g(x^2) = f(g(x)).\,$ Given an initial ansatz of $\, g(x) = O(x)\,$ we use the previous equation to solve for the power series coefficients one at a time with the result $$ g(x) = \frac13 x + \frac1{18} x^2 + \frac{11}{216} x^3 + \frac{7}{324} x^4 + \frac{389}{10368} x^5 + O(x^6). $$ Now if $\,a_0 = g(z)\,$ then $\, a_n = g(z^{2^n})\,$ which is quadratic convergence to the fixed point $0.$ It is unlikely that there is a closed form for $\,g(x).\,$ A similar result holds for the other fixed points.

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