I can achieve $t=2$, $n=21$, $|\mathcal{F}|=7$. Every pair from $\mathcal{F}$ is defeated by some other element on $11$ of the $21$ coordinates. For future answer's sake, I propose that we report these parameters as $(t,n,c,p) = (2,21,7,11/21)$. If I am not mistaken, this is the first $(2,n,c,p)$ anyone has reported with $p>1/2$.

Take $G$ to be the $21$ element sub group of $S_7$ given by maps $\mathbb{Z}/7 \to \mathbb{Z}/7$ of the form $x \mapsto ax+b$, with $a \in \{ 1,2,4 \}$.
Let $G$ act on $\mathbb{R}^7$ by permuting coordinates.
Let $\vec{v} \in \mathbb{R}^7$ have strictly increasing coordinates and let $A$ be the $7 \times 21$ matrix whose columns are the $G$ orbit of $\vec{v}$.

$G$ acts transitively on unordered pairs of distinct elements in $\mathbb{Z}/7$. So it is enough to check the claim for one unordered pair of coordiantes: Say $\{ 1,2 \}$. I find that, in $11$ of the $21$ columns of $A$, the $3$rd entry is greater than the first or second.

I tried this idea with some other permutation groups that are weakly $2$-transitive but not strictly $2$ transitive, but it didn't work for any of the others. (The trouble with strictly $2$-transitive is that for any pair $f$, $g$ in $\mathcal{F}$, we wind up with $f$ and $g$ tied, and $\max(g,h)$ can only be better than $g$, so $f$ never beats $\max(f,g)$.)

For the sake of more amusing conversations, I'll share the story I have in my head: This problem is about Arrow's voting paradox for $t$-faced candidates. Consider $n$ voters and $c$ candidates; the $i$-th coordinate of the $j$-th element of $\mathcal{F}$ is how well voter $i$ likes candidate $j$. If $t=1$, we are just pointing out that there may be no Condorcet winner. To imagine $t=2$, consider a $2$-faced candidate: when talking to voters who prefer $f$'s positions, she says what $f$ would say and, when talking to voters who prefer $g$'s positions, she repeats $g$'s slogans. The question is whether we can find a situation where every $2$-faced candidate loses to some sincere candidate (and by an overwhelming margin.)