Is the Triangle Finding decision problem in $coNTIME(\tilde{O}(n^2))$?
Update: Sadly, it seems that my initial idea (see below) was incorrect, but it led to some fruitful discussion in the comments.  As a result, the question is still open.
Please let me know if you have any ideas.  :)

Initial Idea: One way to solve Triangle Finding is to find all pairs of vertices that are connected by a path of length 2.  Then, you check if there is an edge between any of these pairs.
A standard way to compute all pairs of vertices that are connected by a path of length 2 is to square the adjacency matrix of the graph.  We can perform Boolean matrix multiplication for the squaring.  All of the 1's in the result will represent pairs of vertices that are connected by a path of length 2.
With nondeterminism, we can compute Boolean matrix multiplication of two n by n matrices in $\tilde{O}(n^2)$ time by guessing the witness for Boolean matrix multiplication (the concept of witness for Boolean matrix multiplication has been discussed in [1] https://doi.org/10.1007/s00453-012-9742-3 and [2] https://doi.org/10.1109/SFCS.1992.267748).  Finally, we check in $O(n^2)$ time if all of the pairs of vertices do not form edges.  &lt;- This paragraph doesn't quite work.