measurable linear functionals are also continuous on separable Banach spaces?

If "measurable" means "Borel" or even "Baire measurable", then this is true. It's a special case of a more general result that any Borel homomorphism of Polish groups is continuous. See for instance Kechris, Classical Descriptive Set Theory, Theorem 9.10.

Also, you can drop the word "separable". Suppose $X$ is an arbitrary Banach space and $f : X \to \mathbb{R}$ is a Borel linear functional. To show $f$ is continuous, it suffices to show that for any sequence $x_n \to 0$, we have $f(x_n) \to 0$. But if we let $X_0$ be the closed linear span of $\{x_n\}$, then $X_0$ is a separable Banach space and the restriction of $f$ to $X_0$ is Borel. So $f$ is continuous on $X_0$, meaning $f(x_n) \to 0$ as desired.

If "measurable" means "measurable with respect to a particular Borel measure $\mu$", then this is not true. It could be that $X$ contains a dense subspace $X_0$ of full measure. In that case, we can use Zorn's lemma to choose a linear functional $f$ which is 0 on $X_0$ and nonzero elsewhere. Since $f = 0$ $\mu$-almost everywhere, it is $\mu$-measurable, but not continuous.

One can also come up with examples where $f$ is $\mu$-measurable and discontinuous, and every linear functional which is $\mu$-a.e. equal to $f$ is also discontinuous. This happens, in particular, for Gaussian measures $\mu$; see Bogachev, Gaussian Measures, Theorem 3.7.6, for an equivalent construction. (Of course, as a consequence of measurability, there will be a Borel function $g$ with $f=g$ a.e.; but then $g$ will not be linear!)

You do not even need seperability. Proposition 1.2.29 in the book Barrelled Locally Convex Spaces of Bonet and Perez Carreras says that every Borel measurable linear map from a Baire locally convex space to a locally convex space is continuous. This result is due to Laurent Schwartz and it is related to his more famous Borelian graph theorem (linear maps from a separable Banach space into Souslin locally convex spaces with borelian graph are continuous).