It is equivalent to AC.

Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial function on $A$ for which $p(a)\in a$ for every $a\in\text{dom}(p)$. We place the forcing order on $\P$, so that $q\leq p$ if $q$ extends $p$ to a larger domain, or equivalently, $p=q\upharpoonright\text{dom}(p)$. In particular, being lower in the order means having more information, larger domain, and so on. The empty function is at the top, the largest element of $\P$. Let us also add an object $\bot$ to $\P$ below all others.

The motivating idea is that $\P$ is the forcing notion that adds a choice function for $A$, augmented with $\bot$.

This is a nontrivial bounded lattice, because any two partial functions $p$, $q$ have a least upper bound $p\vee q$, which is their common part as functions, and a greatest lower bound, which is their union $p\cup q$ if they are compatible as functions, and otherwise $\bot$.

I assume that ultrafilters for you cannot be the whole lattice (since otherwise the ultrafilter assertion would become trivialized). Every proper filter in $\P$, I claim, gives rise to a unifying limit partial choice function, the union of the all the functions in the filter, since the filter cannot contain $\bot$ and so all elements of it must be compatible as functions. Furthermore, the limit function arising in this way from an ultrafilter must be totally defined on $A$, since otherwise we could extend it by defining the choice function on one more set $a\in A$ in the collection.

So from an ultrafilter in $\P$ we get a choice function on $A$.

Let me add a note about distributivity, since the lattice $\P$ is not generally a distributive lattice, one for which $p\vee(q\wedge r)=(p\vee q)\wedge (p\vee r)$. The reason is that it could be that $q$ and $r$ are incompatible, which would make $q\wedge r=\bot$ and consequently the LHS would be $p$, but if $q$ and $r$ differ from $p$ on some point $a\in\text{dom}(p)$, then the RHS will be strictly above $p$, lacking that point $a$ in its domain.

So a question remains, I suppose, about the strength of the ultrafilter lemma for distributive lattices. [**Update** Keith Kearnes posted an answer with references to Herrlich & Klimosky, showing that it is also equivalent to AC with nontrivial bounded distributive lattices.]

boundedlattices. Your notation $(L, \leq, \wedge, \vee)$ suggests that you are not making this assumption. Otherwise, it's not true that each filter extends to a maximal filter. For example, the real line with the standard order is a lattice which has no maximal filter at all. $\endgroup$3more comments