For every Bernoulli random variable $X$ with parameter $p$ there exists some Poisson random variable $Y$ with parameter $p$ such that $$P(X\ne Y)\leqslant p(1-\mathrm e^{-p})\leqslant p^2.$$ Thus, for every binomial random variable $S$ with parameter $(n,p)$ there exists some Poisson random variable $Z$ with parameter $np$ such that $$P(S\ne Z)\leqslant np^2.
$$
In particular, for every $k$,
$f(n)=P(S\geqslant k)$ and $\hat f(n)=P(Z\geqslant k)$ are such that $|f(n)-\hat f(n)|\leqslant np^2$.
Furthermore, $\hat f(n)=\dfrac{(np)^k}{k!}g(n)$ where
$$
g(n)=\mathrm e^{-np}\sum_{i\geqslant k}(np)^{i-k}\frac{k!}{(k+i)!},
$$
hence
$$
\mathrm e^{-np}\leqslant g(n)\leqslant\sum_{i\geqslant k}(np)^{i-k}=\frac1{1-np}.
$$
These bounds are nonasymptotic and valid for every $(n,p,k)$ such that $np\lt1$. Do these imply the asymptotics you are interested in?
For the approximation result mentioned in this answer, see R. J. Serfling, Some Elementary Results on Poisson Approximation in a Sequence of Bernoulli Trials (1976), freely available as preprint M374 on this Tech Reports page.
Edit: An improvement of the upper bound of $g(n)$, which might help the OP, is
$$
g(n)\leqslant\mathrm e^{-np}\sum_{i\geqslant k}\frac{(np)^{i-k}}{(k+1)^{i-k}}=\frac{\mathrm e^{-np}}{1-\frac{np}{k+1}}.
$$
Once the idea is understood, one can play with these bounds, and with their variants, for example, for every $(n,p,k)$ such that $np\lt1$,
$$
1-np\leqslant g(n)\leqslant1-c(k)np,
$$
where $c(k)\to1$ when $k\to\infty$.
Best Answer
As a complement to Ragib Zaman fine answer (and Henry's useful comment see for example Farmer and Leth's 'An asymptotic formula for powers of binomial coefficients' which contains, about a closed form for your sum, "..it has only recently been shown that no such formula can exist") let's add a more precise asymptotic expansion (this is conjectured only...) :
$$\frac{\sum_{k=0}^n \binom{n}{k}^3}{\frac{2^{3n+1} }{\sqrt{3} \pi n}}=1 -\frac 1{3n}+\frac 1{3^3n^2}+\frac 1{3^4n^3}+\frac 1{3^5n^4}+\frac {11}{3^7 n^5}+\frac {49}{3^9 n^6}-\frac {317}{3^9 n^7}-\frac{2797}{3^{10} n^8}-\frac{61741}{3^{13} n^9}+\operatorname{O}\left(\frac 1{n^{10}}\right)$$