The similarity between the entropy power inequality and the Brunn-Minkowski inequality is not directly related to convexity - after all, Brunn-Minkowski can be generalised to bounded open sets that are not necessarily convex. (However, many of the proofs of both inequalities use ideas from convexity theory, of course.)

Taking the microstate (i.e. Boltzmann) interpretation of entropy, one can heuristically view the entropy power inequality as a high-dimensional "99%" analogue of Brunn-Minkowski, in which one considers the "99%-sumset" $A \stackrel{99\%}{+} B$ of two high-dimensional sets A and B (this is a somewhat vaguely defined concept, but roughly speaking it is the bulk of the support of two random variables distributed on A and B respectively) rather than the "100%-sumset" A+B. However, thanks to the concentration of measure phenomenon, the 99%-sumset is considerably smaller than the 100%-sumset in high dimensions, leading to the additional exponent of 2 in the entropy power inequality. For instance, consider in high dimensions two balls $B(0,R)$, $B(0,r)$ of radii $R,r$ respectively. Their 100%-sumset is $B(0,R+r)$ of course. But if one takes a random element of $B(0,R)$ and adds it to a random element of $B(0,r)$, the sum concentrates in a much smaller ball - asymptotically, $B(0,\sqrt{R^2+r^2})$ (in fact it concentrates to the boundary of this ball). This boils down to the basic fact that pairs of vectors in high dimensions are typically almost orthogonal to each other. So one morally has

$$ B(0,R) \stackrel{99\%}{+} B(0,r) = B(0,\sqrt{R^2+r^2}) \qquad (1)$$

in the high-dimensional limit.

Anyway, the EPI can be thought of as a rigorous formulation of a ``99% Brunn-Minkowski inequality''

$$ |A \stackrel{99\%}{+} B|^{1/N} \geq \sqrt{ (|A|^{1/N})^2 + (|B|^{1/N})^2 } \qquad (2)$$

in the high-dimensional limit $N \to \infty$ (with $A,B \subset \mathbb{R}^N$ varying appropriately with $N$), which is of course consistent with (1). To see this interpretation, one observes from the microstate interpretation of entropy that if $X$ is a continuous random variable on $\mathbb{R}^n$ with a nice distribution function (e.g. $C^\infty_c$), and $M$ is large, then taking $M$ independent copies $X_1,\dots,X_M$ of $X$ gives a random vector $X^{\otimes M} := (X_1,\dots,X_M)$ in $\mathbb{R}^{N}$ for $N := Mn$ which (by the asymptotic equipartition property) is concentrated in a subset of $\mathbb{R}^N$ of measure $e^{M (H(X)+o(1))}$ in the limit $M \to \infty$ (this is a nice calculation using Stirling's formula; it may help to first work out the case when the probability distribution of $X$ is a simple function rather than a test function). Similarly, if $Y$ is another random variable independent of $X$, then $Y^{\otimes M}$ will be concentrated in a set of measure $e^{M(H(Y)+o(1))}$, while $(X+Y)^{\otimes M}$ is concentrated in a set of measure $e^{M(H(X+Y)+o(1))}$. EPI is then morally a consequence of (2) in the limit $M \to \infty$.

Despite the significant differences between the $99\%$-sumset and $100\%$-sumset in high dimensions, it is still good to think of these concepts as being closely analogous, so that almost any sumset inequality should have an entropy counterpart and vice versa (although in most cases we do not have a direct logical implication between the sumset inequality and the entropy inequality; instead, the inequalities typically have analogous, but not completely identical, proofs). See e.g. this recent paper of Kontoyiannis and Madiman (and the references therein) for some instances of this.

EDIT: Of course, by bounding the $99\%$-sumset by the $100\%$-sumset one can get *some* connection between the two types of inequalities, but usually one gets an inferior estimate when one uses this approach (it completely ignores the effect of concentration of measure), so this is not the "right" way to relate sumset inequalities with their entropy counterparts. For instance, by directly applying the EPI to uniform random variables on $A,B \subset \mathbb{R}^n$ and then using Jensen's inequality, one only gets a weak form
$$ |A+B|^{1/n} \geq \sqrt{ (|A|^{1/n})^2 + (|B|^{1/n})^2 }$$
of the Brunn-Minkowski inequality (compare with (2)). The problem here, of course, is that the sum of two uniformly distributed independent random variables is almost never uniformly distributed.