The sequence grows too rapidly and I doubt that one can get meaningful asymptotics for the largest prime factor. However, one can show that the largest prime factor is bounded by $(n\sqrt{n})^{2^{\pi(n)}}$ where $\pi(n)$ is the number of primes up to $n$, and since $\pi(n) \sim n/\log n$, this implies a stronger bound than what you wanted.
To see this, put $\alpha =1+ \sqrt{2} + \ldots + \sqrt{n}$ and let $\epsilon=(\epsilon_1,\ldots,\epsilon_n)$ be any choice of signs. Set $\epsilon(\alpha)= \sum_{j=1}^{n} \epsilon_j \sqrt{j}$. Note that $\alpha$ and all the $2^n$ values $\epsilon(\alpha)$ all belong to the field $K={\Bbb Q}(\sqrt{p}, 2\le p\le n)$ which has degree $2^{\pi(n)}$ and Galois group $H \simeq ({\Bbb Z}/2)^{\pi(n)}$. Note that the choice of all possible signs $\epsilon$ forms a group $G \simeq ({\Bbb Z}/2)^n$, and $H$ may be thought of as a subgroup of $G$ in the obvious way. Then
$$
A_n = \prod_{\epsilon \in G/H} N_K(\epsilon(\alpha)),
$$
where the product is over the $2^{n-\pi(n)}$ cosets $G/H$, and $N_K$ is the norm in $K$ (the product over all the elements of $H$). Note that all the $N_K(\epsilon(\alpha))$ are non-zero integers. Further all these norms are bounded by $(n\sqrt{n})^{2^{\pi(n)}}$ (there are $2^{\pi(n)}$ conjugates and each is bounded in size by $\sqrt{1} +\ldots + \sqrt{n} \le n\sqrt{n}$).
Thus $A_n$ can be written as the product of $2^{n-\pi(n)}$ integers each of size at most $(n\sqrt{n})^{2^{\pi(n)}}$, and therefore its largest prime factor is no more than that.
