Quantum evasiveness conjecture?
Since the quantum query complexity typically denotes the bounded-error quantum query complexity, there's some ambiguity. A more precise question could be: "What is the quantum query complexity to decide nontrivial monotone graph properties with probability at least $2/3$?".

A natural example in this context is the monotone property of having a single edge. Since Grover search is optimal for this problem, its bounded-error quantum query complexity seems like the natural candidate.



update (2020 May 5):

Since the monotone property of connectivity has quantum query complexity $\Theta(n^{3/2})$ (see here), and the property of having a single edge has quantum query complexity $\Theta(n)$, there is actually no hope for a simple function $f(n)$ only dependent on $n$ to fully characterize the quantum query complexity of monotone graph properties (even up to constants). This is in contrast to what is conjectured for the classical deterministic query complexity.

Alternatively, one could ask about "the minimum possible quantum query complexity of a nontrivial monotone graph property" (as formulated in the paper by Aaronson, Ben-David, Kothari, and Tal). In this case, a simple function of $n$ might still be possible, and the complexity of having a single edge again seems like a natural candidate.



update 2 (2020 May 5):

Answering the updated question, a possible candidate for the additional property would be the (worst-case) certificate size $c$ of the property. A corresponding candidate for the query complexity would then be $\Theta(\sqrt{n^2 c})$, which describes the quantum query complexity when using Grover search on a graph that is empty apart from a single certificate (i.e., a minimal graph having the property). This would be consistent with the examples of having a single edge ($c = 1$) and connectivity ($c = n-1$).

Here's some other consistent examples:


Nonplanarity. Certificate size $c \in \Theta(n)$ (consider the graph obtained by replacing all edges of a $K_5$ graph by paths of length $\Omega(n)$). The quantum query complexity is $\Theta(n^{3/2}) = \Theta(\sqrt{n^2c})$ (link).
Having more than $\binom{n}{2}/2$ edges. Certificate size $\binom{n}{2}/2+1$. The quantum query complexity is $\Theta(n^{3/2}) = \Theta(\sqrt{n^2c})$ (link).
Triangle finding. Has certificate size $c = 3$. The current best upper bound is $\tilde{O}(n^{5/4})$ (Le Gall), but the best lower bound is only $\Omega(n) = \Omega(\sqrt{n^2 c})$.$^1$


$^1$At the end of his QIP talk (link), Le Gall does mention that he would be "surprised" if the quantum query complexity were $\tilde O(n)$.