The density will always be $0$.

Using the sieve of Eratosthenes and the Chebotarev density theorem we can prove that for any positive irreducible polynomial $F\in\mathbb{Z}[X]$, $$\#\left\{ n\leq x:\ F(n)\text{ is prime}\right\} \ll_F\frac{x}{(\log\log x)^{1-o(1)}}.$$ This can be improved to $\ll_F\frac{x}{\log x}$ by using either the fundamental lemma of the Sieve, or the Selberg Sieve.

*Proof:* Lets sieve out by $P(z)=\prod_{p\leq z}p$. Define $$\mathcal{A}=\left\{ F(n):\ n\leq x\right\},$$
and $$S(\mathcal{A},z)=\left|\left\{ a\in\mathcal{A}:\ \gcd(a,P(z))=1\right\} \right|.$$
Then $$\#\left\{ n\leq x:\ F(n)\text{ is prime}\right\} \leq z+S(\mathcal{A},z).$$
Set $\mathcal{A}_{d}=\left\{ a\in\mathcal{A}:\ a\equiv0\text{ mod }d\right\}$. Then since $$\sum_{d|n}\mu(d)=\begin{cases}
1 & \text{if }n=1\\
0 & \text{otherwise}
\end{cases}$$
we may write $$S(\mathcal{A},z)=\sum_{\begin{array}{c}
a\in\mathcal{A}\\
(a,P(z))=1
\end{array}}1=\sum_{a\in\mathcal{A}}\sum_{d|a,\ d|P(z)}\mu(d)=\sum_{d|P(z)}\mu(d)\sum_{\begin{array}{c}
a\in\mathcal{A}\\
d|a
\end{array}}1=\sum_{d|P(z)}\mu(d)|\mathcal{A}_{d}|.$$
Now let $$v_{F}(d)=\left|\left\{ m\in\mathbb{Z}/d\mathbb{Z}:\ F(m)\equiv0\ (\text{mod}\ d)\right\} \right|.$$
Then $$|\mathcal{A}_{d}|=v_{F}(d)\left(\frac{x}{d}+O(1)\right)=x\frac{v_{F}(d)}{d}+O(v_{F}(d)),$$
and so $$S\left(\mathcal{A},z\right)=x\sum_{d|P(z)}\mu(d)\frac{v_{F}(d)}{d}+O\left(\sum_{d|P(z)}v_{F}(d)\right).$$
Since $v_{F}(d)\leq(\deg F)^{\omega(n)},$
we have that $$S\left(\mathcal{A},z\right)\leq x\prod_{p\leq z}\left(1-\frac{v_{F}(p)}{p}\right)+O\left((2\deg F)^{\pi(z)}\right).$$
Now, by the Chebotarev density theorem $$\frac{1}{\pi(x)}\sum_{p\leq x}v_{F}(p)=1+o(1),$$ which implies that $$\prod_{p\leq z}\left(1-\frac{v_{F}(p)}{p}\right)\ll \frac{1}{(\log z)^{1-o(1)}},$$
and so $$S\left(\mathcal{A},x\right)\ll_{F}\frac{x}{(\log z)^{1-o(1)}}+(2\deg F)^{\pi(z)}.$$
Choosing $z=\log x/2,$
we obtain the desired result $$\#\left\{ n\leq x:\ F(n)\text{ is prime}\right\} \ll\frac{x}{(\log\log x)^{1-o(1)}}.$$