Why don't nodes carry as many pointers as possible in a skip list?
In the usual way, where each node carries $i$ pointers with probability $2^{-i}$, the average node has size $O(1)$, so the total space used by the skip list is $O(n)$. If each node carried pointers at every level, then each node would have $\theta(\log n)$ pointers in it, since that is the expected height of a skip list with $n$ nodes. The total space used would thus be $\theta(n \log n)$.

A skip list of one million nodes would thus be using roughly 10 times as much space.