Question: We have that $$(x+y)^n=\sum\_{k=0}^n\binom{n}{k}x^ky^{n-k},$$
but is there such a formula for $(x+y+z)^n$ ?

Answer: 
$$(x+y+z)^n=([x+y]+z)^n=\sum\_{k=0}^n\binom{n}{k}(x+y)^kz^{n-k}=\sum\_{k=0}^n\binom nk\left[\sum\_{j=0}^k\binom{k}{j}x^jy^{k-j}\right]z^{n-k}$$
There is::

$$(x+y+z)^n= \!\!\sum\_{\substack{(i,j,k)\in\mathbf N^3\\i+j+k=n}}\!\frac{n!}{i!\,j!\,k!} x^i y^j z^k.$$
It can be extended to sums of more than $3$ terms (*multinomial formula*):
$$(x\_1+x\_2+\dots+x\_r)^n= \!\!\sum\_{\substack{(i\_1,i\_2,\dots, i\_r)\in\mathbf N^r\\i\_1+i\_2+\dots+i\_r=n}}\!\frac{n!}{i\_1!\,i\_2!\,\dotsm i\_r!}\, x\_1^{i\_1} x\_2^{i\_2}\dotsm \,x\_r^{i\_r}.$$