If, as suggested by Ori Gurel-Gurevich, we sample from uniform distribution on $[0,1]$, then $Z$ will typically be of order $1/\sqrt{n}$. 

A convenient way of generating the points $X_1,\dots,X_n$ in order is letting $W_1,\dots,W_{n+1}$ be independent exponential(1) variables with partial sums $S_k = W_1+\cdots+W_k$, and finally letting $X_k = S_k / S_{n+1}$.
We have $$\left|X_k - \frac{k}{n+1}\right| \leq \frac{\left|S_k-k\right| + \left|S_{n+1} - (n+1)\right|}{n+1},$$
and in particular 
$$\max_{k\leq n} \left|X_k - \mathbb{E}X_k\right| \leq \frac2{n+1}\cdot\max_{k\leq n+1} \left|S_k - k\right|.$$
It is relatively easy to see that $\mathbb{E}\max_{k\leq n} \left|S_k-k\right| = O(\sqrt{n})$ and consequently that $\mathbb{E} \max_{k\leq n} \left|X_k - \mathbb{E}X_k\right| = O(n^{-1/2})$. This is because $S_n-S_k$ is independent of $S_k$. Therefore if $S_k$ deviates wildly from its mean, then with decent probability (namely when $S_n-S_k$ deviates ever so slightly in the same direction), $S_n$ will deviate just as wildly from its mean. More precisely, $$Pr\left(\left|S_n - n\right| > t\right) \geq C\cdot Pr\left(\left|S_k-k\right| > t \ \text{for some $k\leq n$}\right),$$
where $C$ is simply a positive lower bound on the probability that a sum of independent exponentials deviates from its mean in a given direction. Consequently \begin{equation} \mathbb{E} \max_{k\leq n} \left|S_k-k\right| = \int_0^\infty Pr\left(\max_{k\leq n}\left|S_k-k\right| > t\right) \, dt \end{equation} \begin{equation} \leq \frac1C \int_0^\infty Pr\left(\left|S_n-n\right| > t\right)\, dt = \frac1C\cdot \mathbb{E}\left|S_n-n\right| = O(\sqrt{n}). \end{equation}