I think worst case insertion complexity will be $O(n^2)$ if duplicates are not allowed and it will $O(n)$ if duplicates are allowed.
Case 1 - Duplicates allowed :
Lets assume all the elements map to the same slot then for every element we can add it at the front of the chain.So $O(1)$ for each insertion and $O(n)$ for $n$ insertions.
Case 2 - Duplicates not allowed:
Here also all the elements map to the same slot :
Now for $1^{st}$ element we insert it to the slot it maps to.
For $2^{nd}$ element (it also maps to the same slot as said earlier) we need to do lookup whether it is already present or not - So 1 comparison is needed and then it is inserted in $O(1)$ time at the front of the chain if not already present.
For $3^{rd}$ element again do lookup , so 2 comparisons and then it is inserted in $O(1)$ time.
For $4^{th}$ element again do lookup , so 3 comparisons and then it is inserted in $O(1)$ time.
$\cdots$
$\cdots$
$\cdots$
For $n^{th}$ element again do lookup , so $n-1$ comparisons and then it is inserted in $O(1)$ time.
So total time of insertion of $n$ elements = total lookup time + total insert time.
Total lookup time = $1+2+3+...+(n-1)=\frac{n(n-1)}{2}=O(n^2)$
Total insert time $= O(n)$
Total Insertion time for $n$ elements $= O(n^2)+O(n)=O(n^2)$