$t \in (0, 4]$.

First, we show how to construct a good collection of $n$ triangles, each of perimeter greater than $4$. This will show that all $t \le 4$ satisfy the required conditions.

Construct inductively an $(n+2)$-gon $BA_1A_2\dots A_nC$ inscribed in $\omega$ such that $BC$ is a diameter, and $BA_1A_2, BA_2A_3, \dots, BA_{n-1}A_n, BA_nC$ is a good collection of $n$ triangles. For $n=1$, take any triangle $BA_1C$ inscribed in $\omega$ such that $BC$ is a diameter; its perimeter is greater than $2\cdot BC = 4$. To perform the inductive step, assume that the $(n+2)$-gon $BA_1A_2\dots A_nC$ is already constructed. Since $A_nB + A_nC + BC > 4$, one can choose a point $A_{n+1}$ on the small arc $CA_n$, close enough to $C$, so that $A_nB + A_nA_{n+1} + BA_{n+1}$ is still greater than $4$. Thus each of the new triangles $BA_nA_{n+1}$ and $BA_{n+1}C$ has perimeter greater than $4$, which completes the induction step.

We proceed by showing that no $t > 4$ satisfies the conditions of the problem. To this end, we assume that there exists a good collection $T$ of $n$ triangles, each of perimeter greater than $t$, and then bound $n$ from above.

Take $\varepsilon > 0$ such that $t = 4 + 2\varepsilon$.

Claim. There exists a positive constant $\sigma = \sigma(\varepsilon)$ such that any triangle $\Delta$ with perimeter $2s \ge 4 + 2\varepsilon$, inscribed in $\omega$, has area $S(\Delta)$ at least $\sigma$.

Proof. Let $a, b, c$ be the side lengths of $\Delta$. Since $\Delta$ is inscribed in $\omega$, each side has length at most $2$. Therefore, $s - a \ge (2 + \varepsilon) - 2 = \varepsilon$. Similarly, $s - b \ge \varepsilon$ and $s - c \ge \varepsilon$. By Heron's formula, $S(\Delta) = \sqrt{s(s-a)(s-b)(s-c)} \ge \sqrt{(2+\varepsilon)^3\varepsilon}$. Thus we can set $\sigma(\varepsilon) = \sqrt{(2+\varepsilon)^3\varepsilon}$. $\square$

Now we see that the total area $S$ of all triangles from $T$ is at least $n \cdot \sigma(\varepsilon)$. On the other hand, $S$ does not exceed the area of the disk bounded by $\omega$. Thus $n\sigma(\varepsilon) \le \pi$, which means that $n$ is bounded from the above.