Output Voltage of a Current Mode Controlled Flyback Converter
When operating in DCM (and that is what this design does), the voltage transfer is this: -

$$\dfrac{V_{OUT}}{V_{IN}} = N\cdot D\sqrt{\dfrac{R_L}{2\cdot F_{SW}\cdot L_P}}$$

Where: -


\$N\$ is the turns ratio as a fraction i.e. 1/16.67 (0.06) for your case
\$D\$ is the duty cycle percentage
\$R_L\$ is your load resistance on the secondary
\$F_{SW}\$ is your switching frequency i.e. 65 kHz
\$L_P\$ is your primary inductance


The important thing to note is that the voltage transfer is load dependent unlike when operating in CCM.


  How does one make sure this will work for a given output voltage (12V,
  in this case)


Try plugging in the numbers into the above equation but do take account of what your maximum and minimum loading values are.