Yes we have some information using the theory of spacings of order statistics.
Let $X_{(1)}\le \dots \le X_{(n)}$ be the Gaussian sample sorted, the so-called order statistics. What you are asking about is information on $X_{(i+1)}-X_{(i)}$ (may be linked to this post), well more precisely you can rewrite the quantity you are interested in as
$$ \Delta=\min(X_{(i+1)}-X_{(i)},X_{(i)}-X_{(i-1)}) $$
and you are interested about information on this quantity.
Now, Gaussian order statistics are not the easiest ones, and typically to handle this sort of things, the easiest way is to reduce the problem to exponential order statistics because exponential order statistics are well known. To do that, you can consider the function $U=F_{exp}^{-1} \circ \Phi$ where $F_{exp}$ is the cdf of an exponential with parameter $1$ and $\Phi$ the cdf of a standard Gaussian. By the Probability integral transform, you have that if $X\sim \mathcal{N}(0,1)$ then $U(X)\sim Exp(1)$ and because $U$ is increasing,
$$U(X_{(1)})\le \dots \le U(X_{(n)}) $$
Denote $E_i=U(X_i)$, we have that $E_{(1)},\dots,E_{(n)}$ are exponential order statistics ! Now, it is known from order statistics theory that $$i(E_{(i+1)}-E_{(i)})\sim Exp(1)$$
Good, now that we have a handle on this, we come back to the original sample by mean value theorem:
\begin{align*}
X_{(i+1)}-X_{(i)}&=U^{-1}(U(X_{(i+1)}))-U^{-1}(U(X_{(i)}))\\
&\le \left(U(X_{(i+1)})- U(X_{(i)})\right)\sup_{x \in [U(X_{(i)},U(X_{(i+1)}]} U^{-1}(x)\end{align*}
and so on.
For example, this reasoning allow you to get concentration inequalities for $\Delta$, this is the technique that was used for example in the article Concentration inequalities for order statistics by Stephane Boucheron and Maud Thomas.
To go further you can also see wikipedia or the article Spacings Around An Order Statistic by H. N. Nagaraja, Karthik Bharath, Fangyuan Zhang.