I don't know why you think that DC-DC losses are so high. Even when dropping from such a high voltage as 56v (48v bank maximum charge voltage) at a low current you should be able to achieve > 75% efficiency.
As you are playing with something as technical as the Raspberry Pi, I imagine that you can do basic soldering and knock up a circuit on strip board (vero board) or similar?
If so, then why not roll your own SMPS?
National Semiconductor were recently bought by Texas Instruments, but National made a range of SMPS called Simple Switcher which were designed to be easy to design with and make with minimum of tweaking or external components.
http://www.ti.com/lit/ds/symlink/lm2591hv.pdfIf you go to the front page of the TI website, you shall see something called WEBENCH Designer. This would allow you to input the design parameters such as input voltage ragnge, output voltage and current, and then the software would design the circuit for you and suggest suitable components. RS (rswww.com) or Farnell (uk.farnell.com) supply components and can be bought with a credit card.
The above switcher would be suitable and would achieve over 75% efficiently.
Thus your 3.5w Pi load should equate to about 4.66w at the battery. (3.5 / 0.75)
-Tim
PS: The 5V0 rail on the Pi is not regulated at all no matter how you supply the voltage (USB socket or otherwise). Be sure to apply the correct voltage.