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First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$

Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla f(x)|}\,d^{n-1}\sigma(x),$$ to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)\,d^nx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$

So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.

_{Dimensional check: if $x$ has dimension of length $L$ and $\theta_1,\theta_2$ are dimensionless, then the integral should have dimension $L^n$. The area of the $n-2$ dimensional region has dimension $L^{n-2}$, and since the product $|v_1||v_2|$ has dimension of $L^{-2}$ the dimensions match.}

First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$

Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla g(x)|}\,d^{n-1}\sigma(x),$$ once with $g(x)=v_1\cdot x$, $f(x)=\delta(v_2\cdot x)$, then once more with $g(x)=v_2\cdot x$, $f(x)=1$, to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)\,d^nx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$

So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.

_{Dimensional check: if $x$ has dimension of length $L$ and $\theta_1,\theta_2$ are dimensionless, then the integral should have dimension $L^n$. The area of the $n-2$ dimensional region has dimension $L^{n-2}$, and since the product $|v_1||v_2|$ has dimension of $L^{-2}$ the dimensions match.}

First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$

Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla f(x)|}\,d^{n-1}\sigma(x),$$ to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)\,d^nx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$

So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.

First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$

Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla f(x)|}\,d^{n-1}\sigma(x),$$ to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)\,d^nx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$

So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.

_{Dimensional check: if $x$ has dimension of length $L$ and $\theta_1,\theta_2$ are dimensionless, then the integral should have dimension $L^n$. The area of the $n-2$ dimensional region has dimension $L^{n-2}$, and since the product $|v_1||v_2|$ has dimension of $L^{-2}$ the dimensions match.}

First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$

Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla f(x)|}\,d^{n-1}\sigma(x),$$ to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)dx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$

So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.

First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$

Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla f(x)|}\,d^{n-1}\sigma(x),$$ to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)\,d^nx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$

So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.