It is consistent with ZFC that the answer to your question is no. Specifically, I claim, if we assume the continuum hypothesis, then the answer is no, not even for sunflowers of size $n=3$.

**Theorem.** Assume the continuum hypothesis. Then there is a
maximal almost disjoint family with no sunflower of size 3.

The proof will rely on the following lemma.

**Lemma.** If $\cal A$ is a countable almost disjoint family and
$B$ is an infinite set almost disjoint from every member of $\cal
A$, then there is an infinite set $A$ having a different finite
intersection with every element of $\cal A$ and containing
infinitely many elements of $B$.

**Proof.** Enumerate the family ${\cal A}=\{A_0,A_1,A_2,\ldots\}$,
and fix the infinite set $B$ almost disjoint from every $A_n$. We
build the set $A$ in stages. At stage $n$, we will have already
fixed the intersections $A\cap A_k$ for $k<n$, promising to add no
additional elements of $A_k$ to $A$ beyond what has already been
added. Consider $A_n$. By adding some elements of $A_n$ to $A$ not
in any $A_k$ for $k<n$, we can ensure that $A\cap A_n$ is distinct
from the intersections $A\cap A_k$ that we've already fixed. And we
can also add another element of $B$. After doing this, we promise
not to add any more elements from $A_n$. Thus, in countably many
steps, we construct the set $A$ as desired. $\quad\Box$

**Proof of theorem.** By CH we can well order the infinite subsets
of $\mathbb{N}$ in order type $\omega_1$. We shall now form a
maximal almost disjoint family of sets $\langle
A_\alpha\mid\alpha<\omega_1\rangle$, with the further property that
every $A_\alpha$ has a distinct finite intersection with $A_\beta$
for all $\beta<\alpha$. This property will ensure that the family
has no sunflower of size 3.

At stage $\alpha$, consider the least set $B$ in the well order
that is almost disjoint from every $A_\beta$ for $\beta<\alpha$. By
the lemma, there is a set $A_\alpha$ that has a different finite
intersection with every $A_\beta$ for $\beta<\alpha$ and contains
infinitely many elements of $B$.

By construction, we've made an almost disjoint family containing no
sunflowers of size 3. It is a maximal almost disjoint family, since
if $B$ is a set that is almost disjoint from the family, then at
some stage it would have been the least such set, and then we would
have added a set having infinite intersection with it. So this is a
maximal almost disjoint family with no sunflowers of size 3, as
desired. $\quad\Box$

We might consider an almost disjoint family that is maximal with respect to the property of not containing any 3-sunflower (or $\kappa$-sunflower for any cardinal $\kappa$). This suggests a host of new cardinal characteristics, namely, $\frak{a}_\kappa$ is the size of the smallest almost disjoint family that is maximal with respect to the property of not containing any $\kappa$-sunflower. The argument I give shows that $\omega_1\leq\frak{a}_3$, and I guess it is immediate that $\frak{a}_\kappa\leq\frak{a}$ (**Update** this is *not* actually clear). I am less clear on the relation between $\frak{a}_\kappa$ and $\frak{a}_\lambda$ if $\kappa<\lambda$.

**Question.** Can we separate these cardinal characteristics?

For example, can we find a model where $\frak{a}_3<\frak{a}$ or where $\frak{a}_3\neq\frak{a}_4$?

I have posted a question about this at Can we separate the almost-disjointness sunflower numbers?