I think the answer is yes. More generally, suppose $\pi:X\rightarrow Y$ is a surjective map of irreducible schemes over $\mathbb{Q}$ with compactifications $\overline{X},\overline{Y}$ respectively. Pick very ample line bundles $\mathcal{L}_X, \mathcal{L}_Y$ and use them to define heights $h_X,h_Y$. Then there exist $\alpha,\beta>0$ such that for any $y\in Y(\bar{\mathbb{Q}})$ there exists $x$ over it with $h_X(x)\leq \alpha h_Y(y) + \beta$.
I believe these heights are comparable to the canonical heights on a semi-abelian variety, so this should answer your question.

After a finite etale base change the generic point of $Y$ will have a section back to $X$, so by possibly replacing $Y$ by an open subset of it, and making this base change we can assume that $\pi$ has a section $\psi:Y\rightarrow X$. Note that etale base changes won't affect the height by more than a constant. Moreover, replacing $Y$ by an open subset is justified because we can then repeat the argument for the complement of this open set and this process will eventually end by Noetherian induction. We now shrink a bit further to assume that $Y$ is an affine scheme $Y=\rm{spec}(R)$. I claim that one can take $x=\psi(y)$.

Since $\mathcal{L}_Y$ is very ample, we can take a basis of sections $(y_1,\dots,y_n)$ such that $(y_i/y_j)_{i,j}$ generate the fraction field of $R$. Letting $(x_1,\dots,x_m)$ be the corresponding sections of $\mathcal{L}_X$ it follows that there are rational functions $Q_{i,j}(y_1/y_2,\dots,y_1/y_n)$ which are equal to the restriction of $x_i/x_j$ to $\psi(Y)$. The claim follows.