Solution:
Let us call our original $n$ points $V_{1}, V_{2}, \ldots, V_{n}$.
If $A, B$ are two points, then viewers on one side of line $\overleftrightarrow{A B}$ see $A$ to the left of $B$, and viewers on the other side see $B$ to the left of $A$. Therefore, if we draw the $\binom{n}{2}$ lines determined by pairs $\{V_{i}, V_{j}\}$ ($1 \leq i < j \leq n$), then different "views" of $V_{1}, V_{2}, \ldots, V_{n}$ (from outside their convex hull) are in one-to-one correspondence with the regions formed outside the convex hull by these $\binom{n}{2}$ lines. These regions are what we will now count. Our strategy is to count all regions, then subtract those regions that are inside the convex hull of $V_{1}, V_{2}, \ldots, V_{n}$.

We begin by stating and proving a general lemma:

Lemma. Let $K$ be a convex region of the plane. Let $r$ lines pass through the interior of $K$, making a total of $m$ intersections in the interior of $K$. Suppose further that no three of the lines meet at one of these intersections. Then $K$ is divided by the lines into $1 + r + m$ regions.

Proof of Lemma. Number the lines $\ell_{1}, \ell_{2}, \ldots, \ell_{r}$. We imagine that we draw the lines one at a time, in this order. Suppose $m_{i}$ is the number of points in the interior of $K$ where $\ell_{i}$ intersects $\ell_{1}, \ell_{2}, \ldots, \ell_{i-1}$. Then at the stage when we draw $\ell_{i}$, it passes through $m_{i} + 1$ existing regions, dividing each of them into two regions and thereby increasing the number of regions by $m_{i} + 1$. Since there is initially one region ($K$ itself), the final number of regions after all $r$ lines are drawn is
$$1 + (m_{1} + 1) + (m_{2} + 1) + (m_{3} + 1) + \cdots + (m_{r} + 1).$$
But observe that $m_{1} + m_{2} + \cdots + m_{r} = m$, since each intersection point in the interior of $K$ is counted exactly once on each side of the equation. Thus the final number of regions is $1 + r + m$.

By the lemma, if $\binom{n}{2}$ lines are in general position (no two parallel, no three concurrent), then they divide the plane into $1 + \binom{n}{2} + \binom{n}{2}$ regions.

However, our $\binom{n}{2}$ lines aren't in general position; $n-1$ lines meet at each of our original points $V_{i}$. Each $V_{i}$ is surrounded by $2(n-1)$ regions, but if we nudged each line by a tiny amount so as to separate all their pairwise intersections, then these $2(n-1)$ regions would become $1 + n + \binom{n-1}{2}$ regions (by the lemma). Accounting for this, the number of regions formed by our $\binom{n}{2}$ lines is
$$1 + \binom{n}{2} + \binom{n}{2} + n\left[2(n-1) - (1 + n + \binom{n-1}{2})\right].$$

Finally, we subtract the regions inside the convex hull of $V_{1}, V_{2}, \ldots, V_{n}$. The number of lines passing through the convex hull (i.e., diagonals, not sides) is $\binom{n}{2} - n$. Every set of four points $\{V_{i}, V_{j}, V_{k}, V_{m}\}$ ($1 \leq i < j < k < m \leq n$) determines a unique intersection inside the convex hull, so there are $\binom{n}{4}$ such intersections. Thus by the lemma, the convex hull is cut into $1 + \binom{n}{2} - n + \binom{n}{4}$ regions. Subtracting this from our total count of regions in the plane, we conclude that the number of regions outside the convex hull (which is our final answer) is
$$1 + \binom{n}{2} + \binom{n}{2} + n\left[2(n-1) - (1 + n + \binom{n-1}{2})\right] - \left[1 + \binom{n}{2} - n + \binom{n}{4}\right]$$
This may be simplified to
$$\frac{1}{12} n(n-1)\left(n^{2} - 5n + 18\right).$$