What is the probability that a rational prime remains prime in $\mathbb Z[i,\sqrt{-3}]$? Using Chebotarev's density theorem, asymptotically, what is the probability that a rational prime remains prime in $\mathbb Z[i, \sqrt{-3}]$?
If you're referring to $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$, the ring of algebraic integers of $\mathbb{Q}(\zeta_{12})$ (see its page in the LMFDB), then the answer is that no prime from $\mathbb{Z}$ remains prime in $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$.

The intermediate fields are $\mathbb{Q}(i)$, $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$. For a prime from $\mathbb{Z}$ to be prime in both $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$, it must be $p \equiv 5 \bmod 12$. But then this means that $p \equiv 1 \bmod 4$ and you know what that means for $\mathbb{Z}[i]$...

P.S. A Naiade discovered the density theorem twelve centuries before Chebotarev.
Based on Euler results and Chebotarev's density theorem, we have that, asymptotically:

i) One half of all rational primes split in $Z[i]$, and these are the primes $p = 1$ mod $4$. Let's call this set $Split(Z[i])$ and,

ii) One half of all rational primes split in $Z[\sqrt{-3}]$, and these are the primes $p = 1$ mod $3$. Let's call this set $Split(Z[\sqrt{-3}])$

Therefore, the rational primes that split in $Z[i,\sqrt{-3}]$ are the union of $Split(Z[i])$ and $Split(Z[\sqrt{-3}])$. 
We have then, by set rules, that:

$Split(Z[i,\sqrt{-3}]) = Union$ ($Split(Z[i])$,$Split(Z[\sqrt{-3}])$) := {primes| $p = 1$ mod $4$ OR $p = 1$ mod $3$}.

And the rational primes that remain inert in this ring $Inert(Z[i,\sqrt{-3}])]$ are in the complement set $C[Split(Z[i,\sqrt{-3}])]$, and by set rules: 

$Inert(Z[i,\sqrt{-3}])]$ ⊊ $C[Split(Z[i,\sqrt{-3}])]  = C[Union (Split(Z[i]),Split(Z[\sqrt{-3}]))] :=$ {primes | NOT ( $p = 1$ mod $4$ OR $p = 1$ mod $3$) } = {primes| NOT ($p = 1$ mod $4)$ AND NOT ($p = 1$ mod $3$) } = {2,3,primes|  $p = 3$ mod $4$ AND $p = 2$ mod $3$ }

Finally, by Chebotarev's density theorem and probability rules, asymptotically, the probability $P$ that a prime $p = 3$ mod $4$ AND $p = 2$ mod $3$ is equal to the product of the probabilities for each condition:

$P$ (prime $p = 3$ mod $4$ AND $p = 2$ mod $3$) = $P$ (prime $p = 3$ mod $4$) * $P$ (prime $p = 2$ mod $3$) = $1/2 * 1/2 = 1/4$

Since the primes that ramify in $(Z[i,\sqrt{-3}])$ form a finite set, we have that $P(p$ is inert in $Z[i,\sqrt{-3}]$)=$P(p$ doesn't split in $Z[i,\sqrt{-3}]$), therefore $1/4$ of all rational primes remain prime in $Z[i,\sqrt{-3}]$