I'm not completely sure this is what you're looking for, but after playing a bit with the suggested inequality, I think I have found a combinatorial way of thinking about it. In fact, we shall prove the following inequality of which the inequality we are interested in is a special case (in fact, they are equivalent):
Inequality. Let $m,n\in\mathbb{N}$ and let $k_1,k_2,\cdots k_n\in\mathbb{N}$ (yes, this is the same $n$) be natural numbers such that $k_1+k_2+\cdots+k_n=mn$. Then the following inequality holds: $$m^n \geq k_1k_2\cdots k_n\qquad(*)$$
(Taking $m=x_1+x_2+\cdots+x_n$ and $k_i = nx_i$ yields the inequality in the question.)
We shall proceed by a constructiong a sequence of injections. The following lemma is easy to see. (It is basically just the combinatorial interpretation of the inequality $k l\leq (k+1)(l-1)$ which holds for natural numbers $k<l$.)
Lemma. Let $k,l\in\mathbb{N}$ be numbers such that $k<l$. Let $A,B$ be sets such that $|A|=k$ and $|B|=l$ and let $b\in B$. Then there is an injection $\phi:A\times B\to (A\cup\lbrace b\rbrace)\times(B\setminus\lbrace b\rbrace)$.
Proof. Since $k<l$, we have $k\leq l-1$, which means that there is an injection $\psi:A\to B\setminus\lbrace b\rbrace$, so we can define $\phi$ by $$\phi(x,y)=\begin{cases}(x,y);&\textrm{if }y\neq b,\\(b,\psi(x));&\textrm{if }y=b.\end{cases}$$ It is straightforward to verify that this is indeed an injection, so we are done.
From this lemma we get the following proposition, which is the inequality $(*)$ interpreted combinatorially.
Proposition. Let $m,n\in\mathbb{N}$ and let $A_1,A_2,\cdots,A_n$ be disjoint finite sets such that $\sum_{i=1}^n|A_n|=nm$. Let $B$ be a set such that $|B|=m$. Then there is an injection $\phi:A_1\times A_2\times\cdots\times A_n\to B^n$. ($B^n$ denotes $n$-tuples of elements from $B$, as usual.)
Proof. If $|A_i|=m$ for each $i$, there is not much to prove. Otherwise, there are indices $i_1$ and $i_2$ such that $|A_{i_1}|<m$ and $|A_{i_2}|>m$. Now choose some $c\in A_{i_2}$ and define a new partition of $\bigcup_{i=1}^n A_i$ as follows: $A_i^{(1)} = A_i$ for $i\neq i_1,i\neq i_2$ and $A_{i_2}^{(1)}=A_{i_2}\setminus\lbrace c\rbrace$, $A_{i_1}^{(1)}=A_{i_1}\cup\lbrace c\rbrace$. The above lemma allows us to construct an injection $\phi_1:A_1\times A_2\times\cdots\times A_n\to A_1^{(1)}\times A_2^{(1)}\times\cdots\times A_n^{(1)}$.
If now $|A_i^{(1)}|=m$ for each $i$, we are done, otherwise, repeat the same procedure, to obtain a partition $(A_{i}^{(2)})_i$ and an injection $\phi_2:A_1^{(1)}\times A_2^{(1)}\times\cdots\times A_n^{(1)}\to A_1^{(2)}\times A_2^{(2)}\times\cdots\times A_n^{(2)}$.
The procedure will terminate after finitely many steps, since at each step $j$ the sum $\sum_{i=1}^{n}|k_i^{(j)}-m|$ decreases so it will eventually reach zero at some step $r$. (Here we define $k_i^{(j)}=|A_i^{(j)}|$.) But this implies $k_i^{(r)} = m$ for each $i$, so $|A_i^{(r)}|=|B|$ for each $i$, so there is a bijection $\psi:A_1^{(r)}\times A_2^{(r)}\times\cdots\times A_n^{(r)}\to B^n$.
This means we have found a sequence of injections: $$A_1\times A_2\times\cdots\times A_n\overset{\phi_1}{\to}A_1^{(1)}\times A_2^{(1)}\times\cdots\times A_n^{(1)}\overset{\phi_2}{\to}\cdots\overset{\phi_r}{\to}A_1^{(r)}\times A_2^{(r)}\times\cdots\times A_n^{(r)}\overset{\psi}{\to}B^n$$ and the proof is complete.
Conclusion. We have constructed a sequence of injections that proves $(*)$, thus proving the A-G inequality in a combinatorial fashion. The combinatorial meaning of the proof is basically something like this: suppose we have a set $A$ and we partition it into $n$ non-empty parts. Then the number of ways to choose a single element from each set of the partition will be the greatest when the partition is into sets of equal size.
Best Answer
This seems to be deceptively cumbersome, needing a tight bound on the LHS while symmetrisation. The following is from my old notes, unfortunately no way to attribute it correctly....
First we need a well known inequality $4(x+y+z)^3 \geqslant 27(x^2y+y^2z+z^2x+xyz)$, which follows from AM-GM, as WLOG we may assume $y$ is in between $x, z$: $$\frac{\frac{x+z}2+y+\frac{x+z}2}3\geqslant \sqrt[3]{\frac{x+z}2\cdot y \cdot \frac{x+z}2}$$ $$\implies \frac4{27}(x+y+z)^3 = x^2y+y^2z+z^2x+xyz - z(y-x)(y-z) \\\geqslant (x^2y+y^2z+z^2x+xyz)$$
Using the above with $x = \frac{a}b, y=\frac{b}c, z = \frac{c}a$, we get $$\left(\frac{a}b+\frac{b}c+\frac{c}a \right)^3 \geqslant \frac{27}4\left( \frac{a^3+b^3+c^3}{abc}+1\right)$$
Hence to prove the original inequality, it is enough to show instead the symmetric $$\frac{a^3+b^3+c^3}{abc} +1\geqslant 108 \frac{(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{1}$$
$$\iff \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+4\geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{2}$$
Using the obvious $(ab+bc+ca)^2 \geqslant 3abc(a+b+c)$, $$\frac{a+b+c}{abc} \geqslant \frac{3(a+b+c)^2}{(ab+bc+ac)^2}$$ and hence, it is enough to show $$\frac{3(a+b+c)^2(a^2+b^2+c^2-ab-bc-ac)}{(ab+bc+ac)^2} + 4 \geqslant \frac{108(a^2+b^2+c^2)^3}{(a+b+c)^6} \tag{3}$$ Setting $t=\dfrac{3(a^2+b^2+c^2)}{(a+b+c)^2}\, \in [1, 3)$ we need to equivalently show $$\frac{54(t-1)}{(3-t)^2}+4\geqslant 4t^3 \tag{4}$$ which is $(t-1)(9-6t-8t^2+10t^3-2t^4)\geqslant 0$ and hence holds true as $$9-6t-8t^2+10t^3-2t^4=2(3+3t-t^2)(t-1)^2+3>0$$