Iosif Pinelis provided a very nice answer, however, I would like to provide a more comprehensive answer to this question. I think the title is a bit misleading because we do not actually need the order statistics to satisfy any kind of large deviation principle, I would rather call it a limiting result if you let $n\rightarrow \infty$ (Then Hoeffding limiting theorem [1] will kick in and you can derive associated distributional properties for the limiting Gaussian distribution, say study the extremes [2]); and a concentration bound if you want $n$ fixed.
As for this question and OP's "PS" comment, I wanted to point our that a more general result can be obtained from Efron-Stein inequality, by Boucheron [3], and it is sharp:
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[3] **Theorem 2.9 (Exponential Efron-Stein inequality).**
Let $X_{1},\cdots,X_{n}$ be independently distributed according to $F$, let $X_{(1)}\geq\cdots\geq X_{(n)}$ be the order statistics and let $\Delta_k=X_{(k)}-X_{(k+1)}$ be the k-th spacing. Let $V_{k}=k\Delta_{k}^{2}$ denote the Efron-Stein estimate of the variance of $X_{(k)}$ (for k = 1, . . . , n/2). If $F$ has a non-decreasing hazard rate $h$, then for 1\leq k\leq n/2$,
$$Var[X_{(k)}]\leq\boldsymbol{E}V_{k}\leq\frac{2}{k}\boldsymbol{E}\left[\left(\frac{1}{h(X_{(k+1)})}\right)^{2}\right]$$
for $\lambda\geq0,1\leq k\leq n/2$
$$\log\boldsymbol{E}\exp\left[\lambda\left(X_{(k)}-\boldsymbol{E}X_{(k)}\right)\right]\leq\lambda\frac{k}{2}\boldsymbol{E}\Delta_{k}\left(\exp\left(\lambda\Delta_{k}-1\right)\right)=\frac{\lambda k}{2}\boldsymbol{E}\left[\sqrt{\frac{V_{k}}{k}}\left(e^{\lambda\sqrt{\frac{V_{k}}{k}}}-1\right)\right].$$
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With this result we can actually apply the OP's argument
$$P(\sup_{1 \leq i \leq n} \Delta^{(i)} \geq x) \leq \sum_{i=1..n} P( \Delta^{(i)} \geq x)\leq \frac{1}{x^2}\sum_{i=1..n} Var(X_{(i)})$$
on a *wider class of densities* with a mild assumption on Stein estimates, which is known to behave well and well studied. The resulting inequality will still be sharp because every inequality is **sharp**.
One more comment on Iosif's answer. The reason why "uniformly bounded" condition (say even the density is not compactly supported, Iosif's arguments still hold with a slight modification, but as long as "$f$ is uniformly bounded from zero" does not hold, his arguments collapse as noted.) cannot be dropped is exactly the reason why I commented in the beginning this is not a "LDP" type result, which does not assume such a condition.
**Reference**
[1]Randles, Ronald H., and Douglas A. Wolfe. "Introduction to the theory of nonparametric statistics." Introduction to the theory of nonparametric statistics, by Randles, Ronald H.; Wolfe, Douglas A. New York: Wiley, c1979. Wiley series in probability and mathematical statistics (1979).
[2]DasGupta, Anirban, S. N. Lahiri, and Jordan Stoyanov. "Sharp fixed n bounds and asymptotic expansions for the mean and the median of a Gaussian sample maximum, and applications to the Donoho–Jin model." Statistical Methodology 20 (2014): 40-62.
[3]Boucheron, Stéphane, and Maud Thomas. "Concentration inequalities for order statistics." Electronic Communications in Probability 17 (2012).
[4]http://www.stat.purdue.edu/~dasgupta/orderstats.pdf