$\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\QQ}{\mathbb Q}\newcommand{\ZZ}{\mathbb Z}$The polynomial is irreducible over $\mathbb C$ for all $n>3$. This has been confirmed for $n=4$ and $n=5$ by some of the comments, so we may henceforth assume $n>5$. The proof outline is as follows:
- Reduce the question of reducibility over $\CC$ to reducibility over $\FF_p$ for particular primes $p$. (Note that we use reducibility over $\FF_p$, not its algebraic closure.)
- Show that reducibility over $\FF_p$ implies reducibility of $x^n-t$ for many values of $t\in\FF_p$, by substituting particular values and showing that we can recover irreducibility of the polynomial of interest.
- Show that, in enough cases, these values of $t$ comprise all of $\FF_p$.
Lemma 1. Suppose $f\in\ZZ[x_1,\dots,x_n]$ is homogeneous and reducible over $\CC$, and let $g\in\ZZ[y]$ have no double roots in $\CC$. There exist infinitely many primes $p$ for which $f\in\FF_p[x_1,\dots,x_n]$ is reducible and $g\in\FF_p[y]$ splits completely.
Proof. We first need to express the idea that "$f$ is reducible" as a bunch of polynomial equations, and this is where we use the homogeneity of $f$. Since factors of homogeneous polynomials are homogeneous, $f$ is reducible if and only if can be written as the product $f_1f_2$ for some polynomials $f_1$ and $f_2$ of degrees $r$ and $s$, respectively. Suppose $f$ is reducible over $\CC$ with these degrees. Let variables $w_1,\dots,w_M$ and $z_1,\dots,z_N$ represent the coefficients of arbitrary degree $r$ and $s$, respectively, homogeneous polynomials. Letting $x^{\alpha_1},\dots,x^{\alpha_M}$ and $x^{\beta_1},\dots,x^{\beta_N}$ be the corresponding monomials (each represents the product $x_1^{a_1}\cdots x_n^{a_n}$ for some nonnegative integers $a_1,\dots,a_n$), a solution to
$$\left(\sum_{i=1}^M w_ix^{\alpha_i}\right)\left(\sum_{j=1}^N z_jx^{\beta_j}\right)=f(x_1,\dots,x_n)$$
with $w_1,\dots,w_M,z_1,\dots,z_N\in k$ for a field $k$ implies reducibility of $f$ over that field (and we know that such a solution exists with $k=\CC$). Expanding and equating coefficients, we get some polynomials $h_1,\dots,h_K\in\ZZ[w,z]$ so that $f$ is reducible if these polynomials have a common zero.
We know these polynomials have a common zero in $\CC$; let one such zero be $(a_1',\dots,a_M',b_1',\dots,b_N')$. Then the ideal $I$ generated by $h_1,\dots,h_K$ in $\CC[w,z]$ is contained in the maximal ideal $(w_1-a_1',\dots,w_M-a_M',z_1-b_1',\dots,z_N-b_N')$ and is thus proper; in particular, there exist no polynomials $j_1,\dots,j_K\in\CC[w,z]$ for which $h_1j_1+\cdots+h_Kj_K=1$. This means there exist no such polynomials in $\overline\QQ[w,z]$, and as a result the ideal $(h_1,\dots,h_K)\subset\overline\QQ[w,z]$ is proper. This means it is contained in some maximal ideal $\mathfrak m$ of $\overline\QQ[w,z]$; by Hilbert's Nullstellensatz such an ideal must be $(w_1-a_1,\dots,w_M-a_M,z_1-b_1,\dots,z_N-b_N)$ for some $(a_1,\dots,a_M,b_1,\dots,b_N)\in\overline\QQ^{M+N}$. This means that $f$ is reducible over $\overline\QQ$.
We now convert this factorization into one over $\FF_p$ for infinitely many primes $p$. Let $L$ be the number field generated by the union of $\{a_1,\dots,a_M,b_1,\dots,b_N\}$ and the set of roots $\{v_1,\dots,v_k\}$ of $g$ in $\overline\QQ$. By the primitive element theorem we can write $L=\QQ(\alpha)$ for some $\alpha\in L$, whence there exist polynomials $p_1,\dots,p_M,q_1,\dots,q_N,r_1,\dots,r_k\in\QQ[x]$ for which $p_i(\alpha)=a_i$, $q_i(\alpha)=b_i$, and $r_i(\alpha)=v_i$. Let $t\in\ZZ[x]$ be the minimal polynomial of $\alpha$. Note that, since $L\cong \QQ(x)/(t)$,
$$t(x)\mid h_j\big(p_1(x),p_2(x),\dots,p_M(x),q_1(x),\dots,q_N(x)\big)$$
for every $1\leq j\leq K$ and $t(x)\mid g(r_i(x))$ for every $1\leq i\leq k$ (both of these statements hold in $\ZZ[x]$). By an elementary result of Schur
there exist infinitely many primes modulo which $t$ has a root. Take any such prime $\ell$; we claim that $f$ is reducible and and that $g$ splits completely modulo all but finitely many such primes. Indeed, let $\beta\in\FF_\ell$ be a root of $f$ in $\FF_\ell$, and define $a_i^{(\ell)}=p_i(\beta)$, $b_j^{(\ell)}=q_j(\beta)$, and $v_j^{(\ell)}=r_i(\beta)$. We have for each $1\leq j\leq K$ that
$$h_j\big(a_1^{(\ell)},\dots,a_M^{(\ell)},b_1^{(\ell)},\dots,b_N^{(\ell)}\big)=h_j\big(p_1(\beta),\dots,p_M(\beta),q_1(\beta),\dots,q_N(\beta)\big)=t(\beta)u_j(\beta)=0$$
for some polynomial $u_j\in\ZZ[x]$, and for each $1\leq i\leq k$ that $g(v_i^{(\ell)})=g(r_i(\beta))=t(\beta)v_i(\beta)=0$ for some $v_i\in\ZZ[x]$. This implies by the definition of the $h_j$ that, over $\FF_\ell$,
$$\left(\sum_{i=1}^M a_i^{(\ell)}x^{\alpha_i}\right)\left(\sum_{j=1}^N b_j^{(\ell)}x^{\beta_j}\right)=f(x_1,\dots,x_n),$$
so $f$ is reducible over $\FF_\ell$. Also, the values $v_i^{(\ell)}$ are each roots of $g$; since the $v_i$ are distinct, the polynomials $r_i$ are also distinct, and so modulo only finitely many primes does $r_i(\beta)=r_j(\beta)$ (such primes divide the resultant of $t$ and $r_i-r_j$, which is a constant since $t$ is irreducible). This means that $g$ has exactly the roots $v_i^{(\ell)}$ for large enough $\ell$, as desired. $\square$
Corollary 2. If $f=x_1^n+\cdots+x_n^n-nx_1\cdots x_n$ is reducible over $\CC$, there exist infinitely many primes $p\equiv 1\pmod n$ for which $f$ is reducible over $\FF_p$.
Proof. Let $g=\Phi_n(y)$ be the $n$th cyclotomic polynomial, which splits completely in $\FF_p$ if and only if $n\mid p-1$ (i.e. if and only if there are exactly $n$ $n$th roots of unity in $\FF_p^\times$). Applying Lemma 1 to $f$ and $g$ gives the result.
Let $p\equiv 1\pmod n$ be a prime, and let $B\subset\FF_p^\times$ be the unique subgroup of index $n$, i.e. $B=\{x^n: x\in\FF_p^\times\}$. Let $A=\{0\}\cup B\subset\FF_p$. Given a set $S$ inside an abelian group and a positive integer $m$, let $mS=\{s_1+s_2+\cdots+s_m\colon s_1,\dots,s_m\in S\}$. We will need the following somewhat technical result, mostly related to additive combinatorics.
Lemma 3. Let $n>5$. For all sufficiently large $p\equiv 1\pmod n$, $(n-2)A=\FF_p$.
Proof. Note that, since $A$ is permuted by multiplication by elements of $B$, $mA$ is as well for any $m\geq 1$, and so $mA$ consists of $0$ and some multiplicative cosets of $B$. In particular, $|mA|\equiv 1\pmod{|B|}$. By the
Cauchy-Davenport theorem
, we have
$$|2A|\geq 2\big(|A|-1\big)+1=2|B|+1$$
with equality if and only if $A$ forms an arithmetic progression $\{a+rd\colon r\in\{0,1,\dots,|A|-1\}\}$ in $\FF_p$.
Suppose first that $A$ does not form an arithmetic progression. Then $|2A|>2|B|+1$. This implies $|2A|\geq 3|B|+1$, which means
$$|4A|=|2(2A)|\geq \max(2\big(|2A|-1\big),p)\geq \max(6|B|+1,p).$$
This implies, using repeated applications of Cauchy--Davenport,
\begin{align*}
|(n-2)A|
&=\big|4A+\underbrace{A+A+\cdots+A}_{n-6\text{ copies}}\big|\\
&\geq \max\big(|4A|+(n-6)(|A|-1),p\big)\\
&\geq \max(6|B|+1+(n-6)|B|,p\big)=\max(n|B|+1,p)=p.
\end{align*}
This means that $(n-2)A=\FF_p$.
Now, suppose $A$ does form an arithmetic progression; we will show $p=n+1$. Suppose not; such a progression must be of the form
$$\{-sd,-(s-1)d,\dots,-d,0,d,2d,\dots,rd\}$$
for some nonnegative integers $r$ and $s$ with $r+s=|B|$, since $0\in A$. If $r>1$, then $d\in B$ implies that
$$B=\{-s,-(s-1),\dots,-1,1,2,\dots,r\}$$
since $x\in B$ if and only if $xd\in B$. Since $B$ is a subgroup, $2r$ must be in $B$, which means $p\mid 2r-j$ for some $j\in\{-s,\dots,r\}$, and so $p$ has a multiple in the interval $[r,2r+s]$. In particular, $p<2r+s\leq 2(r+s)=2|B|$, a contradiction since $|B|=(p-1)/n$.
On the other hand, if $r\leq 1$, then $s>0$ (since $p>n+1$), and $-d\in B$, implying $B\subset \{-1,1,2,\dots,s\}$. The same argument now applies; $2s\in B$, meaning that $p$ is at most $2|B|-1$, again a contradiction. $\square$
We now use our lemmas to prove the desired result. Suppose that $f=x_1^n+\cdots+x_n^n-nx_1\cdots x_n$ is reducible over $\CC$, and let $p\equiv 1\pmod n$ be a large prime for which $f$ is reducible over $\FF_p$, guaranteed to exist by Corollary 2.
By Lemma 3, there exist $a_1,\dots,a_{n-2}\in A$ for which $t=-(a_1+\cdots+a_{n-2})$ is a primitive root. For each $1\leq i\leq n-2$, let $b_i\in\FF_p$ be so that $b_i^n=a_i$. Define
$$f_1(x)=f(x,0,b_1,\dots,b_{n-2})=x^n+\sum_{i=0}^{n-2}b_i^n-0=x^n-t.$$
We first claim that $f_1$ is irreducible. Indeed, if $x^n-t$ has a degree $d$ divisor, then it has roots in $\FF_{p^d}$, so it suffices to show that $x^n-t$ has no roots in $\FF_{p^d}$ for any $1\leq d<n$. If there is such a root, then $t$ is in the subgroup of index $n$ of $\FF_{p^d}^\times$, and so this subgroup contains the subgroup $\FF_p^\times$ of index $\frac{p^d-1}{p-1}$ in $\FF_{p^d}^\times$. This implies $n\mid\frac{p^d-1}{p-1}$. Since $p\equiv 1\pmod n$,
$$\frac{p^d-1}{p-1}=\sum_{i=0}^{d-1}p^i\equiv \sum_{i=0}^{d-1}1=d\pmod{n},$$
so $d$ must be at least $n$, as desired.
Now, suppose that $f=gh$ with $g,h\in\FF_p[x_1,\dots,x_n]$; we'll show that one of them is constant, contradicting the reducibility of $f$ over $\FF_p$. Defining
$$g_1(x)=g(x,0,b_1,\dots,b_{n-2})\text{ and }h_1(x)=h(x,0,b_1,\dots,b_{n-2}),$$
we see that $f_1=g_1h_1$, so either $g_1$ or $h_1$ is constant; without loss of generality let it be $g_1$. Then $h_1$ contains a scalar multiple of $x^n$ as a monomial, and so $h$ contains a scalar multiple of $x_1^n$ a monomial. Since factors of homogeneous polynomials are homogeneous, $\deg h=n$ implies $\deg g=0$, which means $g$ is constant. $\square$
