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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 $$X_i=(Y_i-Y_{i+1})/2X_i=(Y_i-Y_{i+1})/\sqrt2.$$ Hence, one has also $$(*)=E[\exp(-[(Y_1-Y_2)^++\cdots+(Y_n-Y_{n+1})^+]/2)]. *)=E[\exp(-[(Y_1-Y_2)^++\cdots+(Y_n-Y_{n+1})^+]/\sqrt2)].$$

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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 $$X_i=(Y_i-Y_{i+1})/2.$$ Hence, one has also $$(*)=E[\exp(-[(Y_1-Y_2)^++\cdots+(Y_n-Y_{n+1})^+]/2)].$$

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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, $$(*)\ge\exp(-E[X_1^++\cdots+X_n^+])=\exp(-nE[X_1^+]),$$ hence *)\ge\exp(-E[X_1^++\cdots+X_n^+])=\exp(-nE[X_1^+])=\exp(-n/\sqrt{2\pi}). $$Another (*)\ge\mathrm{e}^{-nc}\qquad\mbox{with}\ c=E[X_1^+]=\frac12+\frac1{\sqrt{2\pi}}.$$ A final 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 $$X_i=(Y_i-Y_{i+1})/2.$$ Hence, one has also $$(*)=E[\exp(-[(Y_1-Y_2)^++\cdots+(Y_n-Y_{n+1})^+]/2)].$$

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