Set family $\mathcal{F}$ such that for all $A,B,C \in \mathcal{F}$ both $A \cap B \not \subseteq C$ and $C \not \subseteq A \cup B $

$f(n)$ grows exponentially in $n$ as may be seen by a simple probabilistic argument: choose $M$ subsets $A_1,\dots,A_M$ at random ($M$ to be specified later) independently. Then for given indices $i,j,k$ the probability that $A_i\cap A_j\subset A_k$ is $(7/8)^n$ (there should be no element which does not belong to $A_k$, but belongs to $A_i$ and to $A_j$), and the probability that $A_k\subset A_i\cup A_j$ is the same. So, if $2M(M-1)(M-2)(7/8)^n<1$, with positive probability there is no such an event. This may be further improved by applying Lovasz Local Lemma, since these events are not much dependent (each event depend on $O(M^2)$ other events only).

Just by the fact that your family needs to be both strongly union-free and intersection-free you can improve the upper-bound. First notice that a family $\mathcal{F} \subseteq 2^{[n]}$ is strongly union-free, iff $\overline{\mathcal{F}}= \{[n] \setminus A|A \in \mathcal{F}\}$ is strongly-intersection free. Let's denote by $f_k(n)$ The size of the largest strongly union- free k-uniform family $\mathcal{F}_k \subseteq {[n] \choose k}$, and by $g_k$ the same for strongly intersection-free. Then $g_k(n)=f_{n-k}(n)$. If you demand that your family is both strongly union-free and intersection-free, then the number of sets in the family in the $k$th layer is at most $\min(f_k(n),f_{n-k}(n))$. So the size of a family $\mathcal{F}$ fulfilling your constraints is at most $n\cdot \max_{k \in [n]}\min(f_k(n),f_{n-k}(n))$. Multiplying by $n$ of course becomes negligible as $n$ grows if you just want bounds on the base of the exponent.

If you look at this paper of Coppersmith and Shearer, they actually give a bound for $k$-uniform strongly union-free families, and then see for which $k$ this bound is largest, and multiply by $n$. I haven't done the calculations, but just because now you have the minimum of $f_k(n)$ and $f_{n-k}(n)$, their method should give you a better upper-bound.