This is more of a long comment than an answer. It should be possible
to compute the number of regions and number of bounded regions using
Whitney's theorem for the characteristic polynomial $\chi(t)$ (Theorem
2.4 of <a
href="https://www.cis.upenn.edu/~cis610/sp06stanley.pdf">these
notes</a>), and Zaslavsky's theorem that the number of regions is
$(-1)^d \chi(-1)$, and the number of bounded regions is (in this
situation) $(-1)^d\chi(1)$ (Theorem 2.5 of the previous link). We need more
than the usual definition of "general position." We want the
position to be generic enough for the argument below (generalized to
$d$ dimensions) to hold.

Here is the computation for $d=2$. First, the empty intersection (the
ambient space $\mathbb{R}^2$) contributes $t^2$ to $\chi(t)$. The
${n\choose 2}$ lines will contribute $-{n\choose 2}t$. Now we must
consider all subsets of the lines that intersect in a point $p$. Let
$p$ be one of the original $n$ points. Then ${n-1\choose 2}$ pairs of
lines intersect in $p$, ${n-1\choose 3}$ triple of lines intersect in
$p$, etc., giving a contribution to $\chi(t)$ of
  $$ {n-1\choose 2}-{n-1\choose 3}+{n-1\choose 4} -\cdots = n-2. $$
We have to multiply this by $n$ since there are $n$ choices for
$p$. There are now $3{n\choose 4}$ choices of two lines that don't
intersect in one of the original $n$ points, but they still intersect
by genericity. Thus we get an additional contribution of $3{n\choose
4}$. It follows that
  $$ \chi(t) = t^2-{n\choose 2}t+n(n-2)+3{n\choose 4}. $$
The number of regions is
  $$ \chi(-1) = \frac 18(n-1)(n^3-5n^2+18n-8). $$
The number of bounded regions is
  $$ \chi(1) = \frac 18(n-1)(n-2)(n^2-3n+4). $$
Can someone extend this argument to $d$ dimensions?