The energy density does dilute during inflation: it has to, or else inflation would never end. The energy density is given by $$\rho = \frac{1}{2}\dot{\phi}^2+ V(\phi)$$ where $V(\phi)$ is some potential energy function. As long as $V(\phi)$ is not strictly constant, the inflaton field will evolve according to $$\dot{\rho} = \dot{\phi}\,(\ddot{\phi} + V'(\phi)).$$ If the total energy of the inflaton was conserved, then indeed we'd have $\dot{\rho} = 0$, but if we look at the Klein-Gordon equation for a scalar field in the Friedmann-Robertson-Walker background, $$\ddot{\phi} - 3H\dot{\phi} + V'(\phi) = 0,$$ we see that there is a drag term, proportional to the Hubble parameter, $3H\dot{\phi}$, that draws energy out of the inflaton as it evolves in its potential.
We therefore have $\dot{\rho} = -3H\dot{\phi}^2$, which is nonzero (and negative) as long as $\phi$ is evolving.
EDIT: I should add that the universe can still accelerate even if $\dot{\rho} \neq 0$. To see this, note that $\ddot{a} = a(\dot{H} + H^2) > 0$ which leads to the following bound,
$$\frac{|\dot{H}|}{H^2} < 1.$$
In words, the universe will accelerate if it is vacuum dominated and the field isn't evolving too quickly relative to the expansion rate.