For $n=3$, the polynomial $h_a(x_1,x_2,x_3)$ is irreducible for every $a \geq 1$. Recall that if $h_a = FG$ with $F$ and $G$ non constant then $F$ and $G$ have to be homogenous. By Bézout's theorem, the projective curves $F=0$ and $G=0$ intersect in $\mathbf{P}^2(\mathbf{C})$, which gives a singular point on the curve $h_a=0$. So it suffices to prove that the curve $h_a(x_1,x_2,x_3)=0$ is smooth. Let us prove this by induction on $a$.

We have $$h_a = \sum_{i+j+k=a} x_1^i x_2^j x_3^k$$ so that $$\frac{\partial h_a}{\partial x_1} = \sum_{i+j+k=a-1} (i+1) x_1^i x_2^j x_3^k$$ and similarly for the other partial derivatives. We get $$\frac{\partial h_a}{\partial x_1} +\frac{\partial h_a}{\partial x_2}+\frac{\partial h_a}{\partial x_3} = (a+2) h_{a-1}$$ Now consider $$\frac{\partial h_a}{\partial x_1} - h_{a-1} = \sum_{i+j+k=a-1} ix_1^i x_2^j x_3^k = x_1 \frac{\partial h_{a-1}}{\partial x_1}$$ By symmetry, we also get a formula involving the other partial derivatives of $h_{a-1}$. So if $x=(x_1:x_2:x_3)$ is a singular point of the curve $h_a=0$ then we must have $h_{a-1}(x)=0$ and $x_i \frac{\partial h_{a-1}}{\partial x_i}(x)=0$ for all $i$. By induction, we must have $x_i=0$ for some $i$. Assume for example $x_1=0$. Then we get $$\sum_{j+k=a} x_2^j x_3^k = \sum_{j+k=a-1} x_2^j x_3^k=0$$ which gives a contradiction.

EDIT : prompted by Will Sawin's comment, the argument now works for every $n \geq 3$. Thanks !

The polynomial $h_a(x_1,\ldots,x_n)$ is irreducible for every $a \geq 1$ and $n \geq 3$. 

Recall that if $h_a = FG$ with $F$ and $G$ non constant then $F$ and $G$ have to be homogenous. By Bézout's theorem, the hypersurfaces $F=0$ and $G=0$ intersect in the projective space $\mathbf{P}^{n-1}(\mathbf{C})$ since $n \geq 3$. This gives a singular point on the hypersurface $h_a=0$. So it suffices to prove that $h_a,\frac{\partial h_a}{\partial x_1},\ldots,\frac{\partial h_a}{\partial x_n}$ have no common zero in $\mathbf{C}^n \backslash \{0\}$. This fact is true for every $a \geq 1$ and $n \geq 2$, and we prove this by induction.

For $a=1$ it is easy. For $n=2$ it amounts to the fact that the polynomial $T^a+\cdots+T+1 = (T^{a+1}-1)/(T-1)$ has distinct roots.

In general, we have $$h_a = \sum_{a_1+\cdots+a_n=a} x_1^{a_1} \cdots x_n^{a_n}$$ so that $$\frac{\partial h_a}{\partial x_i} = \sum_{a_1+\cdots+a_n=a-1} (a_i+1) x_1^{a_1} \cdots x_n^{a_n}.$$ Note that $\sum_{i=1}^n \frac{\partial h_a}{\partial x_i} = (a+n-1) h_{a-1}$. Moreover $h_a=x_i h_{a-1}+R$ for some polynomial $R$ not depending on $x_i$, so that $$\frac{\partial h_a}{\partial x_i}=h_{a-1}+x_i \frac{\partial h_{a-1}}{\partial x_i}.$$ If $x=(x_1,\ldots,x_n)$ is a common zero of $h_a$ and all its partial derivatives then $h_{a-1}(x)=0$ and $x_i \frac{\partial h_{a-1}}{\partial x_i}(x)=0$ for all $i$. By induction, we must have $x_i=0$ for some $i$. Assume for example $x_n=0$. Then $(x_1,\ldots,x_{n-1}) \in \mathbf{C}^{n-1}$ provides in fact a common zero of $h_a(x_1,\ldots,x_{n-1})$ and all its partial derivatives, so applying the induction hypothesis for $n-1$ we get $x=0$.