Solution:

The answer is 2167. First of all, 2167 satisfies the required property with $S=2$: indeed
$$\sqrt{2+\sqrt{1+\sqrt{6+\sqrt{7+2}}}}=\sqrt{2+\sqrt{1+\sqrt{6+\sqrt{9}}}}=\sqrt{2+\sqrt{1+\sqrt{9}}}=\sqrt{2+\sqrt{4}}=\sqrt{4}=2$$
Let us now show that there are no integers $n$ between 2016 and 2166 that have this property. We begin by showing that $S$ must be less than 4. Indeed, if for contradiction we had $S \geq 4$, we would have
$$S(S-1)>9 \Rightarrow S^{2}>S+9 \geq S+d \Rightarrow S>\sqrt{d+S}$$
similarly we would have $S^{2}>S+9>\sqrt{d+S}+c \Rightarrow S>\sqrt{c+\sqrt{d+S}}$, and so on, until we reach the contradiction $S>\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d+S}}}}=S$. Observe next that, since $S$ is an integer, all the arguments of the square roots must be perfect squares, and since we are looking for a number between 2016 and 2166 we must have $a=2$ and $b=0$ or $b=1$. Moreover we have $S>1$, because $S>\sqrt{a} \geq 1$, and from $S \leq 3$ we find $d+S \leq 12$, from which $c+\sqrt{d+S}<9+4=13$; in the same way, all the arguments of the square roots are less than or equal to 12. In particular, $\sqrt{12} \geq \sqrt{b+\sqrt{c+\sqrt{d+S}}}=S^{2}-a=S^{2}-2$, from which we obtain $S<3$, that is, $S=2$. We then have $2 \leq d+S \leq 11$, and the only two perfect squares in this interval are 4 and 9, which correspond to $d=2$ and $d=7$ respectively. Let us treat the two cases separately.

d=2. The same situation just seen for $d$ arises again for $c$. If $c=2$ we should have that $b+\sqrt{2+2}$ is a perfect square, but this is true neither for $b=0$ nor for $b=1$. If instead $c=7, b+\sqrt{7+2}$ is a perfect square for $b=1$, which however leads us to $2172>2167$.

d=7. In this case $c+\sqrt{d+S}=c+3$ must be a perfect square between 3 and 12, from which $c=1$ or $c=6$. In the first case again $b+2$ is not a perfect square for $b=0,1$, while for $c=6$ we obtain that $b+3$ must be a square, which happens for $b=1$ and leads to the solution 2167.