How to calculate gain of two cascaded stages low pass filter (passive)?
The transfer function is: $$\small G(s)=\frac{1}{(RC)^2s^2+3RCs+1}$$

Hence \$\omega_n=\frac{1}{RC}=\small10^4\:rad/s\:(=1592\:Hz)\$, and \$\small\zeta=1.5\$, and it can be seen that the DC gain (\$\small s=0\$)  is unity.

Converting this to the frequency domain, using \$ s\rightarrow j\omega\$:

$$\small G(j\omega)=\frac{1}{1-(\omega RC)^2+j3\omega RC}$$

At \$\small 10\:\small Hz\$, \$\small \omega RC=0.00628\$, hence the gain is almost unity and the phase angle is almost zero. At \$\small 1\:\small kHz\$, \$\small \omega RC=0.628\$, giving a gain of \$\small 0.505\$, and phase angle of \$\small \phi=-72^o\$.

So it seems that there's a problem with your experimental set-up. What's the input impedance of the instrument measuring Vout?

Let's do some detective work:

If the input impedance of the instrument were \$\small 3 \: k\Omega\$ resistive, then (i) the gain and phase at DC would be \$\small 0.6\$ and zero, respectively (i.e. same as your results); and (ii) the gain and phase at \$\small 1590 \:\small Hz\$ would be \$\small 0.29\$ and \$\small -79^o\$, which compares with your measurement of \$\small 0.31\$ and \$\small -73^o\$.