Are all Lagrangians translationally invariant?
When you write $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x))$, you're assuming translational invariance. A more general Lagrangian is written $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x),x)$.
Yes they are all translationally invariant as long as they don't depend explicitly on $x^\mu$. As Tong assumes by writing $\mathcal{L}(\phi,\partial_\mu\phi)$.
No, the argument wouldn't go through for Lorentz transformations as they also affect the indices $\mu$ in $\partial_\mu$ and in whatever other field that is not a scalar. That is, $\phi'_{\mu_1\cdots \mu_\ell}(x)$ is not $\phi_{\mu_1\cdots \mu_\ell}(x-\epsilon)$.
The argument wouldn't go through for conformal transformations neither. Both for the reason above and for the fact that if $\epsilon$ depends on $x$ then it does not go through derivatives and so $\partial_\mu\phi$ transforms differently than $\phi$.
That's not necessarily the form of the Lagrangian after a conformal transformation.