This is not a full answer but, first, to point out that things are probably simpler for random than for deterministic $\zeta_i$s. If the $\zeta_i$ are i.i.d. exponential with parameter $1$, for every real numbers $x_i$,
$$
P[\zeta_1\ge x_1,\ldots,\zeta_n\ge x_n]=\exp(-[x_1^++\cdots+x_n^+]),
$$
where $x^+=x$ if $x\ge0$ and $0$ otherwise. Hence,
$$
(*)=P[\zeta_1\ge X_1,\ldots,\zeta_n\ge X_n]=E[\exp(-[X_1^++\cdots+X_n^+])].
$$
A second step (certainly not optimal) is that, the exponential being convex, by Jensen's inequality,
$$
(*)\ge\exp(-E[X_1^++\cdots+X_n^+])=\exp(-nE[X_1^+])=\exp(-n/\sqrt{2\pi}).
$$
Note that the random variables $X_i$ are independent when their indices are at distance at least $2$, hence
$$
(*)\le E[\exp(-X_1^+)]^{\lfloor n/2\rfloor}.
$$
Another (nonconclusive) remark is that a Gaussian random vector $(X_i)_{1\le i\le n}$ with correlation structure as in the post can be realized with the help of an i.i.d. centered reduced Gaussian random vector $(Y_i)_{1\le i\le n+1}$ as 
$$
X_i=(Y_i-Y_{i+1})/\sqrt2.
$$
Hence, one has also
$$
(*)=E[\exp(-[(Y_1-Y_2)^++\cdots+(Y_n-Y_{n+1})^+]/\sqrt2)].
$$