Claim. $n_\min=5$.
Proof: At least one element must be positive, so the system can be rewritten as $$\left\{\begin{array}{l} 1+b_{1}+\cdots+b_{n-1}>0 \\1+b_{1}^{3}+\cdots+b_{n-1}^{3}<0 \\1+b_{1}^{5}+\cdots+b_{n-1}^{5}>0 \end{array}\right.$$ where $b_i=a_{i+1}/a_1$. Intuitively, there should be an infinite number of solutions when $n>4$ as there are more variables than inequalities.
When $n=1$ there is nothing to prove.
When $n=2$, the inequalities $1+b_1>0$ and $1+b_1^3<0$ contradict each other.
When $n=3$, we have $-(1+b_1^5)^{1/5}<b_2<-(1+b_1^3)^{1/3}$ and $-(1+b_1)<b_2<-(1+b_1^3)^{1/3}$. The first double inequality is equivalent to $(1+b_1^5)^3>(1+b_1^3)^5$ which is true only when $b_1\in(-1,0)$; this can be shown by considering the sign of $$f'(b_1)=15b_1^2(b_1+1)^2(2b_1^2-2b_1+1)(2b_1^5-2b_1^4+2b_1^3+1)$$ where $f(b_1)=(1+b_1^3)^5-(1+b_1^5)^3$. The second double inequality is equivalent to $(1+b_1)^3>1+b_1^3$ which is false when $b_1\in(-1,0)$; contradiction.
When $n=4$, we have \begin{align}-(1+b_1^5+b_2^5)^{1/5}<b_3&<-(1+b_1^3+b_2^3)^{1/3}\\-(1+b_1+b_2)<b_3&<-(1+b_1^3+b_2^3)^{1/3}.\end{align} The second double inequality is equivalent to $(1+b_1+b_2)^3>1+b_1^3+b_2^3$, or that $$(b_1+b_2)(b_1+1)(b_2+1)>0\implies\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)\left(c_2^{1/5}+1\right)>0$$ where $c_1=b_1^5$ and $c_2=b_2^5$.
The first double inequality is equivalent to $g(c_1,c_2)=\left(1+c_1^{3/5}+c_2^{3/5}\right)^{5/3}-(1+c_1+c_2)<0$. When $c_2>-1$, we have the criterion $\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)>0$. Now the partial derivative $$g_{c_1}(c_1,c_2)=c_1^{-2/5}\left(1+c_1^{3/5}+c_2^{3/5}\right)^{2/3}-1$$ gives the stationary point $c_1^*=-2^{-5/3}\left(1+c_2^{3/5}\right)^{5/3}$ which we can show is the global minimum by considering second derivatives. In addition, we have $$g(-c_2,c_2)=g(-1,c_2)=0,$$ and $-c_2,-1$ are the only roots since $g_{c_1}$ is strictly decreasing on $(-\infty,c_1^*)$ and increasing on $(c_1^*,\infty)$. Hence $g(c_1,c_2)<0$ only when $c_1\in(\min\{-1,-c_2\},\max\{-1,-c_2\})$, which contradicts the criterion.
When $c_2<-1$, we have the criterion $\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)<0$. But then, in a similar vein to above, $g(c_1,c_2)<0$ only when $c_1\in\Bbb R\setminus(\min\{-1,-c_2\},\max\{-1,-c_2\})$ and we immediately obtain the contradiction. $\square$
Remark 1. When $n=5$, suppose that $b_1=b_3$ and $b_2=b_4$ solve the system. Akin to the proof of the case $n=3$, we have the two inequalities $(1/2+b_1^5)^3>(1/2+b_1^3)^5$ and $(1/2+b_1)^3>1/2+b_1^3$. Numerically, both inequalities hold in the interval $$b_1\in(-0.897,-0.809)\cup(0.309,0.605)$$ which explains the first set of your solutions.
Remark 2. The proof in the case $n=4$ can be greatly reduced in length if we can prove the following generalisation:
If $p\in(0,1)$ is a rational number with an odd numerator and denominator, then the inequality $1+x^p+y^p<(1+x+y)^p$ has solution $(1+x)(1+y)(x+y)<0$ for all such $p$.
This should probably be investigated in a separate question. For the purposes of this problem, the inequality $g(c_1,c_2)<0$ is equivalent to the above with $p=3/5$, where the inequality $(1+c_1)(1+c_2)(c_1+c_2)<0$ would contradict $\left(c_1^{1/5}+c_2^{1/5}\right)\left(c_1^{1/5}+1\right)\left(c_2^{1/5}+1\right)>0$.
