The Newton Method over $\mathbb{R}$ has the property that the precision is doubled (under some continuous differentiable assumption) in each iteration. For the ring $\mathbb{Z}_p$ of $p$-adic integers, we also have the Hensel's lemma, but the precision is increased by 1 in each iteration. The proof about the double precision in the case over $\mathbb{R}$ uses the Taylor expansion and its remainder, see here (Wayback Machine).

In a special case that is finding square root, one can use $x_{n+1} := \frac{1}{2} x_n - \frac{3}{2} x_n^3 $ ( by considering $x^{-2}-c^{-1}$ instead of $x^2-c$) and one can show that the precision is doubled in the case over $\mathbb{Z}_p$.

In general, is the precision is doubled in each iteration in the case over $\mathbb{Z}_p$, under some conditions?

Sorry, I just found that in fact the precision is doubled in $p$-adic case, without further assumptions. I mentioned the precision increased by 1, since this is written in "A first course in $p$-adic Analysis" by Alain M. Robert.

But if one examines the proof, we have $p(\hat{x}) \equiv 0 \ \mathrm{mod} \ p^{n-2k}$. And if $k = 0$, the precision is doubled.

proveswhy the iteration doubles the number of correct digits, but I would say that theexplanationis that the Newton formula $F_f: x \to x - f(x)/f'(x)$ turns the zeros of $f$ into ramified fixed points of the function $F_f(x)$. Basic dynamics then explains why points in a neighborhood of a ramified fixed point end up approaching the fixed point at this rate. In other words, the convergence isn't really a special property of Newton iteration. (I realize this is philosophy, not mathematics.) $\endgroup$ – Joe Silverman May 23 '17 at 17:52