Removing duplicates from unsorted LinkedList
Yes, this looks \$O(n^2)\$: when all elements are unique,
the program will compare all elements with all other elements.

A better solution will be to use a Set to keep track of the already seen items.
You will still have to iterate over all items,
but you won't have to compare to all other elements,
but simply do a lookup in the Set,
which in case of a HashSet will be \$O(1)\$ complexity.
Of course, when all elements are unique,
then this method will use twice as much memory as your original.
That's a trade-off you'll have to consider.

Unfortunately you didn't provide the implementation of LinkedListNode so it's not easy to play with this code.
But looking at it I'm wondering if it correctly removes all duplicates,
for example will it turn 1-&gt;2-&gt;2-&gt;2-&gt;2-&gt;3 into 1-&gt;2-&gt;3.
(Maybe it does, a bit hard to decipher, but I suggest to do a quick test.
And maybe next time include enough code so that reviewers can test run easily.)

UPDATE

As you remarked in your last comment,
another way, without using a buffer / Set is to sort the list first,
and remove the duplicates after.
That would have \$O(n\log n)\$ complexity.
However, keep in mind that the elements will be reordered with that approach (if they were not ordered to begin with).
Depending on your use case, that may be perfectly acceptable, of course.