Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let 
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial.  Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?

In general, if $p_1,\dotsc,p_n$ is any list of integers, then the polynomial
$$f=\prod_{e_1,\dotsc,e_n \in \{\pm 1\}} (x+e_1 \sqrt{p_1}+\dotsc+e_n \sqrt{p_n})$$
has coefficients in $\mathbb{Z}$, which follows by induction from the observation

$p \in \mathbb{Z},\,g \in \mathbb{Z}[x] \Rightarrow g(\sqrt{p}) \cdot g(-\sqrt{p}) \in \mathbb{Z}[x]$.

You can prove this using the automorphism group of $\mathbb{Z}[\sqrt{p}]$. But you can also just verify it via some direct calculation, which has the advantage that you really see why the $\sqrt{p}$-terms vanish and that you may compute $f$ faster.
Clearly, $f$ is monic, has degree $2^n$, and has $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ as a root. So this is quite elementary and for any fixed $n$, you can also compute $f$. What is not so easy to prove is that if $p_1,\dotsc,p_n$ are square-free and pairwise coprime integers, then $f$ is irreducible. Equivalently, the degree of $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ equals $2^n$. You can find the proof here ("square-root-extension.pdf").
For the numbers $p_1,p_2,p_3 = 2,3,5$, we get $f = x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$.
