The following is a ZFC example, due to Michael Hrušák, of a MAD family without sunflowers of cardinality $3$.

Start with the standard AD family $\mathcal{B}=\{B_f:f\in{}^\omega2\}$ of branches through the binary tree $2^{<\omega}$, so $B_f=\{f|n:n\in\omega\}$. Extend $\mathcal{B}$ to a MAD family by adding a family $\mathcal{C}$ that consists of antichains in the three with the additional property that each $C\in\mathcal{C}$ converges to point $b_C$ in ${}^\omega2$ in the sense that for every $n$ the set $\{c\in C:b_C|n\subseteq c\}$ is cofinite in $C$. Every infinite subset of the tree that is almost disjoint from all members of $\mathcal{B}$ contains such a set, so this yields a MAD family.

Next enumerate $\mathcal{C}$ as $\{C_f:f\in{}^\omega2\}$ in a one-to-one fashion and in such a way that $f\neq b_{C_f}$; we write $b_f$ for $b_{C_f}$. Define $A_f=B_f\cup D_f$, where $D_f$ is a co-finite subset of $C_f$ specified as follows: let $k=\min\{n:f(n)\neq b_f(n)\}$, then $D_f=\{c\in C_f:\operatorname{dom}c\ge k+2$ and $c(k)\neq f(k)\}$.

The family $\{A_f:f\in{}^\omega2\}$ is a MAD family without $3$-element sunflowers. Let $f,g,h\in{}^\omega2$ and assume without loss of generality that $k=\min\{n:f(n)\neq g(n)\}$ is larger than or equal to $l=\min\{n:f(n)\neq h(n)\}$ and $m=\min\{n:g(n)\neq h(n)\}$. It follows easily that then in fact one has $l=m<k$.
Let $s$ be the point in $B_f\cap B_g$ whose domain is $l+1$. Then $s$ is not in~$A_h$: it is not in~$B_h$ because $s(l)\neq h(l)$, it is also not in $D_h$, because its direct predecessor is in~$B_h$ and none of the points in $D_h$ have their direct predecessor in $B_h$. It follows that $s\in (A_f\cap A_g)\setminus A_h$, so $\{A_f,A_g,A_h\}$ is not a sunflower.

If one uses the tree $k^{<\omega}$ instead of the binary tree then one create a MAD family with many sunflowers of cardinality $k$ but none of cardinality $k+1$.

The following is a ZFC example, due to Michael Hrušák, of a MAD family without sunflowers of cardinality $3$.

Start with the standard AD family $\mathcal{B}=\{B_f:f\in{}^\omega2\}$ of branches through the binary tree $2^{<\omega}$, so $B_f=\{f|n:n\in\omega\}$. Extend $\mathcal{B}$ to a MAD family by adding a family $\mathcal{C}$ that consists of antichains in the tree with the additional property that each $C\in\mathcal{C}$ converges to point $b_C$ in ${}^\omega2$ in the sense that for every $n$ the set $\{c\in C:b_C|n\subseteq c\}$ is cofinite in $C$. Every infinite subset of the tree that is almost disjoint from all members of $\mathcal{B}$ contains such a set, so this yields a MAD family.

Next enumerate $\mathcal{C}$ as $\{C_f:f\in{}^\omega2\}$ in a one-to-one fashion and in such a way that $f\neq b_{C_f}$; we write $b_f$ for $b_{C_f}$. Define $A_f=B_f\cup D_f$, where $D_f$ is a co-finite subset of $C_f$ specified as follows: let $k=\min\{n:f(n)\neq b_f(n)\}$, then $D_f=\{c\in C_f:\operatorname{dom}c\ge k+2$ and $c(k)\neq f(k)\}$.

The family $\{A_f:f\in{}^\omega2\}$ is a MAD family without $3$-element sunflowers. Let $f,g,h\in{}^\omega2$ and assume without loss of generality that $k=\min\{n:f(n)\neq g(n)\}$ is larger than or equal to $l=\min\{n:f(n)\neq h(n)\}$ and $m=\min\{n:g(n)\neq h(n)\}$. It follows easily that then in fact one has $l=m<k$.
Let $s$ be the point in $B_f\cap B_g$ whose domain is $l+1$. Then $s$ is not in~$A_h$: it is not in~$B_h$ because $s(l)\neq h(l)$, it is also not in $D_h$, because its direct predecessor is in~$B_h$ and none of the points in $D_h$ have their direct predecessor in $B_h$. It follows that $s\in (A_f\cap A_g)\setminus A_h$, so $\{A_f,A_g,A_h\}$ is not a sunflower.

If one uses the tree $k^{<\omega}$ instead of the binary tree then one create a MAD family with many sunflowers of cardinality $k$ but none of cardinality $k+1$.

The following is a ZFC example, due to Michael Hrušák, of a MAD family without sunflowers of cardinality $3$.

Start with the standard AD family $\mathcal{B}=\{B_f:f\in{}^\omega2\}$ of branches through the binary tree $2^{<\omega}$, so $B_f=\{f|n:n\in\omega\}$. Extend $\mathcal{B}$ to a MAD family by adding a family $\mathcal{C}$ that consists of antichains in the three with the additional property that each $C\in\mathcal{C}$ converges to point $b_C$ in ${}^\omega2$ in the sense that for every $n$ the set $\{c\in C:b_C|n\subseteq c\}$ is cofinite in $C$. Every infinite subset of the tree that is almost disjoint from all members of $\mathcal{B}$ contains such a set, so this yields a MAD family.

Next enumerate $\mathcal{C}$ as $\{C_f:f\in{}^\omega2\}$ in a one-to-one fashion and in such a way that $f\neq b_{C_f}$; we write $b_f$ for $B_{C_f}$. Define $A_f=B_f\cup D_f$, where $D_f$ is a co-finite subset of $C_f$ specified as follows: let $k=\min\{n:f(n)\neq b_f(n)\}$, then $D_f=\{c\in C_f:\operatorname{dom}c\ge k+2$ and $c(k)\neq f(k)\}$.

The family $\{A_f:f\in{}^\omega2\}$ is a MAD family without $3$-element sunflowers. Let $f,g,h\in{}^\omega2$ and assume without loss of generality that $k=\min\{n:f(n)\neq g(n)\}$ is larger than or equal to $l=\min\{n:f(n)\neq h(n)\}$ and $m=\min\{n:g(n)\neq h(n)\}$. It follows easily that then in fact one has $l=m<k$.
Let $s$ be the point in $B_f\cap B_g$ whose domain is $l+1$. Then $s$ is not in~$A_h$: it is not in~$B_h$ because $s(l)\neq h(l)$, it is also not in $D_h$, because its direct predecessor is in~$B_h$ and none of the points in $D_h$ have their direct predecessor in $B_h$. It follows that $s\in (A_f\cap A_g)\setminus A_h$, so $\{A_f,A_g,A_h\}$ is not a sunflower.

If one uses the tree $k^{<\omega}$ instead of the binary tree then one create a MAD family with many sunflowers of cardinality $k$ but none of cardinality $k+1$.

The following is a ZFC example, due to Michael Hrušák, of a MAD family without sunflowers of cardinality $3$.

Start with the standard AD family $\mathcal{B}=\{B_f:f\in{}^\omega2\}$ of branches through the binary tree $2^{<\omega}$, so $B_f=\{f|n:n\in\omega\}$. Extend $\mathcal{B}$ to a MAD family by adding a family $\mathcal{C}$ that consists of antichains in the three with the additional property that each $C\in\mathcal{C}$ converges to point $b_C$ in ${}^\omega2$ in the sense that for every $n$ the set $\{c\in C:b_C|n\subseteq c\}$ is cofinite in $C$. Every infinite subset of the tree that is almost disjoint from all members of $\mathcal{B}$ contains such a set, so this yields a MAD family.

Next enumerate $\mathcal{C}$ as $\{C_f:f\in{}^\omega2\}$ in a one-to-one fashion and in such a way that $f\neq b_{C_f}$; we write $b_f$ for $b_{C_f}$. Define $A_f=B_f\cup D_f$, where $D_f$ is a co-finite subset of $C_f$ specified as follows: let $k=\min\{n:f(n)\neq b_f(n)\}$, then $D_f=\{c\in C_f:\operatorname{dom}c\ge k+2$ and $c(k)\neq f(k)\}$.

The family $\{A_f:f\in{}^\omega2\}$ is a MAD family without $3$-element sunflowers. Let $f,g,h\in{}^\omega2$ and assume without loss of generality that $k=\min\{n:f(n)\neq g(n)\}$ is larger than or equal to $l=\min\{n:f(n)\neq h(n)\}$ and $m=\min\{n:g(n)\neq h(n)\}$. It follows easily that then in fact one has $l=m<k$.
Let $s$ be the point in $B_f\cap B_g$ whose domain is $l+1$. Then $s$ is not in~$A_h$: it is not in~$B_h$ because $s(l)\neq h(l)$, it is also not in $D_h$, because its direct predecessor is in~$B_h$ and none of the points in $D_h$ have their direct predecessor in $B_h$. It follows that $s\in (A_f\cap A_g)\setminus A_h$, so $\{A_f,A_g,A_h\}$ is not a sunflower.

If one uses the tree $k^{<\omega}$ instead of the binary tree then one create a MAD family with many sunflowers of cardinality $k$ but none of cardinality $k+1$.