We successively have:
$$\begin{align*} 4 = abc + bcd + cda + dab &= ab(c+d) + cd(a+b) \\ &\le \frac{a^2+b^2}{2} \cdot \sqrt{2(c^2+d^2)} + \frac{c^2+d^2}{2} \cdot \sqrt{2(a^2+b^2)} \\ &= \sqrt{(a^2+b^2)(c^2+d^2)} \cdot \left( \sqrt{\frac{a^2+b^2}{2}} + \sqrt{\frac{c^2+d^2}{2}} \right) \\ &\le \frac{(a^2+b^2)+(c^2+d^2)}{2} \cdot \sqrt{(a^2+b^2)+(c^2+d^2)}. \end{align*}$$

It follows that $4 \le \frac{1}{2} \cdot \sqrt{(a^2 + b^2 + c^2 + d^2)^3}$, i.e. $a^2 + b^2 + c^2 + d^2 \ge 4$.
Equality holds when $a = b = c = d = 1$.

Alternative Solution:

Adding $a^2 + b^2 \ge 2ab$ with the other five similar inequalities and using MacLaurin's inequality we obtain
$$\sum a^2 \ge 4 \left( \frac{\sum ab}{6} \right) \ge 4 \left( \frac{\sum abc}{4} \right)^{2/3} \ge 4,$$
with equality if and only if $a = b = c = d = 1$.