$\mathfrak{ufo}$: An unidentified combinatorial cardinal characteristic of the continuum?

This invariant is known for Boolean algebras in general as pseudo-altitude. That it is $\omega_1$ is proved in van Douwen's chapter in the Boolean algebra handbook 11.1 and 12.7, in the more general form that this holds for any weakly countably complete BA.

It is the case that $\mathfrak{ufo} = \aleph_1$, by considering Shelah's construction of ultrafilters over $\lambda$ with $lcf(\omega, \mathcal{D})= \kappa$ for $\aleph_0 < \kappa \; (\leq 2^{\lambda})$. So for instance, we can let $\kappa = \aleph_1$, and then the construction gives an increasing sequence of $\aleph_1$ many filters on $\lambda$ whose union is an ultrafilter on $\lambda$; the cofinality of the construction is $\aleph_1$. See Classification Theory Ch. VI, namely Theorem 3.12 (page 357/366) and especially Lemma 3.18 (page 360).

We take a large family of independent functions with range $\omega$ and arrange these into $\aleph_1$ many rows. Then we can find an increasing sequence of filters that are maximal with respect to the remaining rows being independent. This give you an ultrafilter ornament as each step fails to be an ultrafilter due to the remaining independent functions, and the union is an ultrafilter. ($\mathcal{G}^*$ is empty in the parlance of Lemma 3.18).

Note this cannot give us a countable sequence of filters as we truly need that the cofinality of the number of rows is uncountable for Lemma 3.18. Thus we do get $\mathfrak{ufo} = \aleph_1$.

A (Slightly) More Detailed Proof

Definition: For a filter $\mathcal{F}$, a set of functions, $\mathcal{G} = \{ f_{\alpha}: \lambda \rightarrow \omega \mid \alpha < \kappa \}$ is independent mod $\mathcal{F}$ if for any finite set of distinct $\{f_1, ... , f_n\} \subset \mathcal{G}$, any finite set $i_1, ... , i_n \in \omega$, and any $F \in \mathcal{F}$: $$\{ j \in \omega \mid f_k(j) = i_k \textrm{for all k} \} \cap F \neq \emptyset. $$

Then it is an old result that there is a family in $\omega^\omega$, of cardinality $2^{\aleph_0}$, that is independent mod $\{ \omega \}$. (For instance, see Kunen Theorem 3.4).

Really, the linked proof must be modified to account for only having a range of $\omega$ for general $\lambda$, but this is easily accomplished by letting the $r$ in the referenced Engelking-Karłowicz proof have range $\omega$. However, it is simple enough to show that there is a family of at least size $\aleph_1$ by a diagonal argument on the finite sets of functions/indices (with respect to the Frechet (cofinite) filter so that we don't end up generating a principle filter). Then we can (with Zorn's) extend this to a maximal family of independent functions and work with that.

Then we can arrange this family into $\aleph_1$ many disjoint pieces; for $\beta < \omega_1$ let $\mathcal{G}_{\beta}$ be the first $\beta$ many pieces. The first claim is then that for each $\beta < \omega_1$, we can build a filter $\mathcal{D}_{\beta}$ using $\mathcal{G}_{\beta}$ that is as large as possible with the constraint that $\mathcal{G} - \mathcal{G}_{\beta}$ is independent mod $\mathcal{D}_{\beta}$. This is done by adding sets of the form $A_n = \{ j \mid f_{\alpha}(j) \geq n \}$. Note our independence property for the functions gives us that each of these can be added without causing an empty intersection somewhere. Then we maximize our filter while keeping the remaining functions still independent (Claim 3.15(3) page 358).

Then finally the second claim is that the union of all of these is maximal with respect to the empty family of functions, i.e. it is an ultrafilter. This is the key Lemma 3.18 with $\mathcal{G}^* = \emptyset$. Part i) just states that we are generating each filter from the family of independent functions in that row; and ii) and iii) say that we are maximizing each of our $\aleph_1$ many steps. Then the $\mathcal{D}^*$ is our desired ultrafilter.

To show that the union is an ultrafilter, it suffices to show that any function from $\lambda$ to $\omega$ is not independent mod the union. This is because if we only had a filter we could build a function that is independent. So consider any such function $f$. If it was in our original $\mathcal{G}$, then it will not be independent mod the union as it is already taken care of at some step. If it was not in $\mathcal{G}$, then it must be dependent, due to the maximality of our family $\mathcal{G}$. So some $i \in \omega$ has $f^{-1}(i)$ with empty intersection with finitely many terms from $\mathcal{G}$. Then removing that index from our function gives us a new function where we can do the same procedure. We do this countably many times (less than cofinality of $\aleph_1$) to get some point where the entire function has already been taken care of in our construction. This is the contrapositive of the statement that a collection that is not an ultrafilter has some countable partition of the underlying set that has some element intersecting the complement of every set in the filter.