If I walk through list and delete every out-of-order element I come across, on average how many elements will be left?
You can compute the average using linearity of expectation.

Let the random variable $X$ denote the number of elements that are retained.  Let $X_i$ be an indicator r.v. that is 1 if the $i$th element is retained, or 0 otherwise.   Then $X = X_1+\dots + X_n$, so

$$\mathbb{E}[X] = \mathbb{E}[X_1] + \dots + \mathbb{E}[X_n] = \Pr[X_1=1] + \dots + \Pr[X_n=1].$$

Now the $i$th element is retained if and only if it is the largest number among the first $i$ elements in the list.  Therefore, $\Pr[X_i=1]=\frac{1}{i}$, so

$$\mathbb{E}[X] = \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \approx \log(n) + \gamma$$

In other words, the average length of the remaining list is $O(\log n)$.

I don't know what the distribution is.

As a heuristic, the expected running time of Drop Sort would satisfy a recurrence something like $T(n) \approx T(n-\log(n)) + O(n)$, which corresponds to $O\left(\frac{n^2}{\log n}\right)$ time.