$\newcommand{\R}{\mathbb{R}}$ Let $U:=\{u\in\R^n\colon u_i=u_{n-i}\ \forall i\}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $\R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $\cdot$ denotes the dot product, your integral is \begin{align} I&:=\int_U du\,\int_{\R^n}dv\,\exp\big(-(\|u\|^2+u\cdot v+\|v\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dv\,\exp\big(-(3\|u\|^2/4+\|v+u/2\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dw\,\exp\big(-(3\|u\|^2/4+\|w\|^2)/2\big) \\ &=(2\pi)^{n/2}\int_U du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}\int_{\R^{n/2}} dt\,\exp\big(-3\|t\|^2/8\big) \\ &=(2\pi)^{n/2}(2\pi)^{n/4}(4/3)^{n/4}. \end{align} The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that \begin{equation} \int_U du\,f(u)=\int_{\R^{n/2}} dt\,f(Tt) \end{equation} for all nonnegative Borel-measurable functions $f\colon U\to\R$, where $T\colon\R^{n/2}\to U$ is a linear isomorphism.

$\newcommand{\R}{\mathbb{R}}$ Let $U:=\{u\in\R^n\colon u_i=u_{n-i}\ \forall i\}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $\R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $\cdot$ denotes the dot product, your integral is \begin{align} I&:=\int_U du\,\int_{\R^n}dv\,\exp\big(-(\|u\|^2+u\cdot v+\|v\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dv\,\exp\big(-(3\|u\|^2/4+\|v+u/2\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dw\,\exp\big(-(3\|u\|^2/4+\|w\|^2)/2\big) \\ &=(2\pi)^{n/2}\int_U du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}\int_{\R^{n/2}} dt\,\exp\big(-3\|t\|^2/8\big) \\ &=(2\pi)^{n/2}(2\pi)^{n/4}(4/3)^{n/4}. \end{align} The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that \begin{equation} \int_U du\,f(u)=\int_{\R^{n/2}} dt\,f(Tt) \end{equation} for all nonnegative Borel-measurable functions $f\colon U\to\R$, where $T\colon\R^{n/2}\to U$ is a linear isometry.

$\newcommand{\R}{\mathbb{R}}$ Let $U:=\{u\in\R^n\colon u_i=u_{n-i}\ \forall i\}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $\R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $\cdot$ denotes the dot product, your integral is \begin{align} I&:=\int_U du\,\int_{\R^n}dv\,\exp\big(-(\|u\|^2+u\cdot v+\|v\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dv\,\exp\big(-(3\|u\|^2/4+\|v+u/2\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dw\,\exp\big(-(3\|u\|^2/4+\|w\|^2)/2\big) \\ &=(2\pi)^{n/2}\int_U du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}\int_{\R^{n/2}} du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}(2\pi)^{n/4}(4/3)^{n/2}; \end{align} The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$.

$\newcommand{\R}{\mathbb{R}}$ Let $U:=\{u\in\R^n\colon u_i=u_{n-i}\ \forall i\}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $\R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $\cdot$ denotes the dot product, your integral is \begin{align} I&:=\int_U du\,\int_{\R^n}dv\,\exp\big(-(\|u\|^2+u\cdot v+\|v\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dv\,\exp\big(-(3\|u\|^2/4+\|v+u/2\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dw\,\exp\big(-(3\|u\|^2/4+\|w\|^2)/2\big) \\ &=(2\pi)^{n/2}\int_U du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}\int_{\R^{n/2}} dt\,\exp\big(-3\|t\|^2/8\big) \\ &=(2\pi)^{n/2}(2\pi)^{n/4}(4/3)^{n/4}. \end{align} The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that \begin{equation} \int_U du\,f(u)=\int_{\R^{n/2}} dt\,f(Tt) \end{equation} for all nonnegative Borel-measurable functions $f\colon U\to\R$, where $T\colon\R^{n/2}\to U$ is a linear isomorphism.