I'm evaluating the sum and I've gone through things below:
Consider $C_{-2}=0,C_{-1}=-1 \\$ where $C_n$ stands for the nth Catalan number $$ S_n=∑_{i=1}^ni^2 C_{i-1} C_{n-i-2}=∑_{i=0}^ni^2 C_{i-1} C_{n-i-2}\\ ∑_{n=0}^∞S_n t^n=∑_{n=0}^∞∑_{i=0}^ni^2 C_{i-1} C_{n-i-2} t^n\\ =∑_{n=0}^∞∑_{i=0}^∞i^2 C_{i-1} C_{n-2} t^{n+i}\\ =(∑_{i=0}^∞C_{i-1}t^i)(∑_{n=0}^∞n^2 C_{n-2} t^n)\\=t(-\sqrt{1-4t})(∑_{n=0}^∞(n+1)^2 C_{n-1}t^n)\\ =t(-\sqrt{1-4t})(∑_{n=0}^∞((n+2)(n+1)-(n+1)) C_{n-1}t^n)\\ =t(-\sqrt{1-4t})(\frac{d^2}{dt^2}∑_{n=0}^∞C_{n-1} t^{n+2}-\frac{d}{dt}∑_{n=0}^∞C_{n-1} t^{n+1})\\ =t(-\sqrt{1-4t})(\frac{d^2}{dt^2}(-t^2 \sqrt{1-4t})-\frac{d}{dt}(-t\sqrt{1-4t}))\\ =t(-\sqrt{1-4t})(\frac{-36 t^2+ 14 t - 1}{(1 - 4 t)^{3/2}}) $$ And I don't know how to do the next step to find the expression of $S_n$ in the form of n.
It is easy to see that the generating function of this sequence of numbers factorizes in two pieces that can be managed using the generating function of the Catalan numbers.
It is well known that if the sequence has the form
$$S_n=\sum_{k=0}^n a_k b_{n-k}$$
then the generating functions of these sequences defined respectively as $S/A/B(x):=\sum_{k=0}^\infty (s/a/b)_n x^n$ are related by
$$S(x)=A(x)B(x)$$
Then, if we consider $a_k=k^2C_{k-1}~,~b_{k}=C_{k-2}$ it remains to calculate $A(x), B(x)$. Define the generating function for the Catalan numbers, which is known in closed form to be:
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}$$
Then we can express $A,B$ in terms of this and its derivatives as follows:
$$A(x)=x (x^2c(x))''-x(xc(x))'~~~,~~~ B(x)=-x+x^2c(x)$$
Then $S$ can be calculated explicitly:
$$S(x)=\frac{x^2(2x-1)}{2}\left(\frac{1}{1-4x}+\frac{1}{(1-4x)^{3/2}}\right)$$
which can be easily reexpanded in series form by noting that
$$\frac{1}{(1-4x)^{3/2}}=\sum_{n=0}^\infty\frac{(n+1)(n+2)}{2}C_{n+1}x^n$$
which after some algebraic manipulations can be rewritten in the form
$$S(x)=-\sum_{n=2}^\infty\frac{(n-1)^2}{2}C_{n-2}x^n-\sum_{n=3}^\infty 4^{n-3}x^n$$
which gives the final form $(S_0=S_1=0,S_2=-1):$
$$S_n=-4^{n-3}-\frac{(n-1)^2}{2}C_{n-2}~~,~~ n\geq 3$$