Firstly notice that $f(1)=0$ so $f(x)=(x+2)g(x)$ over $\mathbb{F}_3$ for some $g$. Moreover, $\text{ord}_{11}3 = 5$, ie $5$ is the smallest exponent $t$ such that $3^t=1 \mod 11$, so $5$ is the lcm of all the degrees of the factors of $f$, hence $f(x)=(x+2)g(x)h(x)$ with $\deg(g)=\deg(h)=5$.
Hence the splitting field of $f(x)$ over $\mathbb{F}_3$ is $L=\mathbb{F}_{3^5}$. Now, is $L$ the smallest extension of $\mathbb{F}_3$ containing all the zeros of $f(x)$? How to prove it?
You are searching for the smallest field extension of $\Bbb F_3$ which contains a primitive $11$-th root of unity.
You're right that the smallest $t$ such that $11$ divides $3^t-1$ (which is the order of the multiplicative group of $\Bbb F_{3^t}$) is $t=5$. Since the group is cyclic it contains all $11$-th roots of unity. That's it.
Add: We have $3^5-1= 11\cdot 22$ and so if $\omega$ is a primitive element of $\Bbb F_{3^5}$, then $\omega^{22}$ is a primitive $11$-th root of unity.