Minimum Value

By AM-GM on each term we have:

$(1+a)(1+b)(1+c)(1+d) \geq 2 \sqrt{a} \times 2 \sqrt{b} \times 2 \sqrt{c} \times 2 \sqrt{d} =16 \sqrt{abcd} =16$

With equality iff $a=b=c=d=1$

By Holder's Inequality,

$(1+a)^{\frac{1}{4}}(1+b)^{\frac{1}{4}}(1+c)^{\frac{1}{4}}(1+d)^{\frac{1}{4}} \geq (abcd)^{\frac{1}{4}}+(1*1*1*1)^{\frac{1}{4}}$

$(1+a)^{\frac{1}{4}}(1+b)^{\frac{1}{4}}(1+c)^{\frac{1}{4}}(1+d)^{\frac{1}{4}} \geq (1)^{\frac{1}{4}}+(1)^{\frac{1}{4}}$

$((1+a)^{\frac{1}{4}}(1+b)^{\frac{1}{4}}(1+c)^{\frac{1}{4}}(1+d)^{\frac{1}{4}})^4 \geq 2^4$

$(1+a)(1+b)(1+c)(1+d) \geq \boxed{16}$

Just assume 1 for all of them. It makes it simple.