Output voltage of this filter
I expect an output of about 1.01 V for your circuit.

Taking KCL at the op-amp's inverting input, you have

$$I({\rm R3}) = I({\rm R2})$$

where both currents are taken as flowing from right to left. Assuming the inverting input is driven to be equal to the non-inverting input by negative feedback and applying Ohm's law, we have 

$$\frac{V_o - 1.65}{390}=\frac{1.65-3.3}{1000}$$

where \$V_o\$ is the voltage at the op-amp's output terminal.

Solving this, \$V_o=1.0065\ {\rm V}\$.

Possibly your 3.3 V source has enough equivalent output resistance to disturb this equation and give the 0.95 V you measured. But in any case, that is much closer to the expected value than 0.625 V is.