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:
[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 $\lambda\geq0,1\leq k\leq n/2$
$$\log\boldsymbol{E}\exp\left[\lambda\left(X_{(k)}-\boldsymbol{E}X_{(k)}\right)\right]\leq\lambda\frac{k}{2}\boldsymbol{E}\Delta_{k}\left(\exp\left(\lambda\Delta_{k}-1\right)\right)=\frac{\lambda k}{2}\boldsymbol{E}\left[\sqrt{\frac{V_{k}}{k}}\left(e^{\lambda\sqrt{\frac{V_{k}}{k}}}-1\right)\right].$$
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 $V_k$, which is known to behave well and well studied. The resulting inequality will still be sharp because every inequality is sharp. And following this line, Iosif's "uniformly bounded below from zero" assumption is no longer needed and hence a more general result can be obtained.
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).