I can only offer a partial solution. Since $$P(\min x_i \le x)=1-(1-F(x))^n$$ with $F$ the $\chi_\nu^2$ CDF, $\min x_i$ has pdf $$n(1-F(x))^{n-1}f(x),$$so the mean you seek is $$\mu:=\int_0^\infty n(1-F(x))^{n-1}xf(x)dx.$$Use integration by parts with $u=x,\,v=-(1-F)^n$ so$$\mu=\int_0^\infty (1-F)^n dx$$(proof that the boundary term vanishes is left as an exercise). But for odd $\nu$ we don't even have an elementary expression for $F$, so I doubt what you're asking for is possible.
Here is something you could want to do if you are provided the function $F:\mathbb{R}^n\rightarrow [0,1]$ such that $F(\mathbf{x})=\mathbb{P}(X_1\leq x_1, \dots , X_n\leq x_n)$.
Observe that you can evaluate the marginal $F(x_1)=\mathbb{P}(X_1\leq x_1)=\lim_{x_2\to \infty}\lim_{x_3\to\infty}\dots\lim_{x_n\to\infty} F(\mathbf{x})$ and the conditional CDF $F(x_2, \dots , x_n \mid x_1) = \mathbb{P}(X_2\leq x_2,\dots, X_n\leq x_n\mid X_1\leq x_1)=F(\mathbf{x})/F(x_1)$. you can apply this strategy recursively to get all the $F(x_1)$, $F(x_2\mid x_1)$, ... , $F(x_n\mid x_1, \dots , x_{n-1})$. Now observe that in the end $F(\mathbf{x})$ is the product of those. You can now generate $p_1\in [0,1]$, then solve as you did $F(x_1)=p_1$, plug this value into $F(x_2\mid x_1)$, generate $p_2$ and find $x_2$, etc...
Does that make sense ?
I will make an EDIT later on how you can probably use newton's method to solve for the $1-D$ case faster, then you can apply it for $n-D$.
EDIT: So we try to generate a sample using the CDF $F$ and PDF $f$, generate $p\in[0,1]$ uniformly, we use the Newton's method to find the roots of $F(x)-p$ (be careful on when it would converge and the speed of convergence). The derivative of this function with respect to $x$ is $f(x)$. you can take a initial guess $x_0$ and then update steps are
$$
x_{n+1}=x_n-\frac{F(x_n)-p}{f(x_n)}
$$
and you can see how you converge using $F(x_n)-p$. You should also know that if the function $f$ is not continuous, for example when the random variable is discrete, then there maybe some problems. To solve this problem, you are supposed to take the smallest of $x$ such that $F(x)=p$.
I would suggest that you take a close look at this, your method may be better than this one for certain cases.
Best Answer
For $d=1$ this is known as Laplace distribution.
You can sample from it by first sampling $R = \|X\|_2$ (which will have a gamma distribution), then sampling $Y$ from a uniform distribution on the unit sphere (e.g. $Y = \frac{Z}{\|Z\|_2}$ where $Z$ has $N(0,1)$ components), and finally setting $X = RY$.