In this case $n=16$ and the point is $x=1$.
I know $\sqrt{x}$ is not differentiable in zero, but this function actually has left derivatives.
Also, I know that $f(x)=(\sqrt{x^2-1}+\sqrt{x-1})^2=x^2+x-2+2\sqrt{(x^2-1)(x-1)}$, so it is enogth to compute the derivative of $2\sqrt{(x^2-1)(x-1)}$.
Also, I tried to find the $n$th derivative of $\sqrt{g(x)}$ but its expresion was too complicated.
HINT:
Since $x\ge 1$, we may write $\sqrt{(x^2-1)(x-1)}=(x-1)\sqrt{x+1}$.
Then, use the General Leibniz Rule for product differentiation
$$\frac{d^n}{dx^n}\left((x-1)\sqrt{x+1}\right)=\sum_{k=0}^n\binom{n}{k}\frac{d^k}{dx^k}(x-1)\frac{d^{n-k}}{dx^{n-k}}(x+1)^{1/2}$$
Finish by recognizing that the only surviving terms occur at $k=0$ and $k=1$.
For convenience take $x = 1+t$, so $x=1$ corresponds to $t=0$.
Thus
$$ \eqalign{f(x) &= \left(\sqrt{(1+t)^2-1} + \sqrt{t}\right)^2\cr
&= t^2 + 3t + 2 + 2 \sqrt{t^2+2t}\sqrt{t}\cr
&= t^2 + 3t + 2 + 2 t \sqrt{t+2}\cr
&= t^2 + 3t + 2 + 2 \sum_{k=0}^\infty \dfrac{\sqrt{2} (-1)^k (2k)! t^{k+1}}{8^k (k!)^2 (1-2k)}
}$$
The $16$'th derivative at $t=0$ comes from the $k=15$ term:
$$ \eqalign{\dfrac{d^{16}}{dt^{16}} \dfrac{2\sqrt{2} (-1)^{15} 30! \; t^{16}}{8^{15} (15!)^2 (-29)} &= \dfrac{2\sqrt{2} (-1)^{15} 30! 16!}{8^{15} (15!)^2 (-29)} \cr &= \frac{213458046676875}{33554432} \sqrt{2}}$$
