Mathematica can actually solve the recursion relation in closed form, $$F_n(n)=-\tfrac{1}{2}(n^2-1)^{-1}\left(\frac{n-1}{n}\right)^n\left[n \left(\frac{n}{n-1}\right)^n \Phi \left(\frac{n}{n-1},1,n+1\right)+n (n+2) \left(\frac{n}{n-1}\right)^n \Phi \left(\frac{n}{n-1},2,n+1\right)-(n-1) \left((n+2) \text{Li}_2\left(\frac{n}{n-1}\right)+3 n+\ln(1-n) +6\right)\right],$$ with $\Phi$ the Lerch transcendent and $\text{Li}_2$ the polylog. The large-$n$ limit then evaluates to $$\lim_{n\rightarrow\infty} F_n(n)=\frac{18+\pi^2}{12 e}.$$
If the initial value is varied, $F_n(0)=p$, the limit becomes $$\lim_{n\rightarrow\infty} F_n(n)=\frac{6p+\pi^2}{12 e}.$$
the Mathematica commands are:
result=f[k]/.RSolve[{f[k]==1/k^2+(1-1/n)*(k+n+1)*f[k-1]/(k+n+2),f[0]==3},f[k],k]/.k->n
Limit[FullSimplify[Normal[Series[result,{n,Infinity,0}]],Assumptions->n>1],n->Infinity]