Adding to comments about breadboarding SMPS... that inductor is a signal inductor - not a power inductor. Whilst it will be 3.3uH, its saturation current will be very low and DCR possibly high enough to bring down efficiency.
Also, a CR2450 won't be able to supply much instantaneous current. Whilst you're after 5v @5mA (= 25mW, so for 60% eff. thats 25/0.6 = ~42mW in, = 14mA input current) the peak current in the inductor will be much higher, which will cause the battery voltage to sage during on-time.
You could scope the inductor current with a 1 ohm resistor in series with the inductor.