What will happen if I drive 2 TEC-12715s and 2 CPU fans off a 12v/30a power supply?
TL;DR:

It's ALWAYS a bad idea to operate outside the current rating of any PSU.

Don't say I didn't warn you.

However, as you will be operating it at 12V, and after looking at the datasheet, your highest current consumption will probably be 12.5A per cell. 12.5A x 2 = 25A, so you'll have 4A headroom after 1A for the fans. It may be safe.

Long-form answer:

If you PSU reaches it current limit, it will start to reduce the voltage supplied. It can easily drop to almost zero in an overcurrent condition.

A good PSU should be protected against this happening and be able to recover itself. OTOH, the modules/devices you connect to it (especially its interfacing circuitry) may not be well protected against an undervoltage condition.

Even if nothing gets permanently damaged, just upset, you may briefly lose functionality while the PSU, peltier cells and fans recover from the overcurrent event. This could be tolerable or not in your application. Maybe your cooling system has enough thermal inertia to keep temperatures low until everything recovers. Or maybe not and you end up with performance glitches as the processor backs off due to temperature rising. That's for you to discover.

However, as you will be operating it at 12V, and after looking at the datasheet, your highest current consumption will probably be 12.5A per cell. 12.5A x 2 = 25A, so you'll have 4A headroom after 1A for the fans. It may be safe.

Anyway, try to be sure that the PSU can supply that much power safely (you may need some degree of forced convection). Otherwise, it's not a good idea.

N.B. - calculation of \$I_{max}\$ at 12V:




Worst case (maximum) current consumption at 12V for which we have data happens when \$T_h=50ºC\$ and \$DT=0ºC\$. Looking up the corresponding values from the curves above we obtain:

$$
\begin{align}
Q_c \approx &amp;\ 150\ W\\
C.O.P.\approx &amp;\ 1
\end{align}
$$

On the other hand, \$C.O.P.=\dfrac{Q_c}{V\cdot I}\$  thus:

$$
I=\frac{Q_c}{V\cdot C.O.P.} \approx \frac{150\ W}{12\ V \cdot 1} = 12.5\ A
$$