Thanks for the clarifications. Unfortunately, there are no nontrivial scalar conformal invariants subject to the restrictions that you have placed on their form, i.e., that the coordinate expression $P[g]$ should be polynomial in $g_{ij}$ and its derivatives plus $\det(g_{ij})^{-1}$. You have to allow more complicated expressions than polynomials if you want scalar conformal invariants.
For example, the lowest order tensorial invariant of a conformal structure in dimension $3$ is the Cotton-York tensor $C[g]$, which is a third-order polynomial invariant that is a section of the bundle
$$T^*M\otimes\Lambda^2(T^*M) = T^*M\otimes TM\otimes\Lambda^3(T^*M)=
\mathrm{End}(TM)\otimes \Lambda^3(T^*M).$$ Unfortunately, the 'trace contraction' $\mathrm{tr}\bigl(C[g])\bigr)$, which is a section of $\Lambda^3(T^*M)$ (and so would be a $3$-form that you could integrate over $M$), vanishes identically.
The trace of the square $C[g]^2$ is a section of $\bigl(\Lambda^3(T^*M)\bigr)^{\otimes2}$ and hence is the square of a density $\sigma_2[g]$ that could be integrated over $M$, but the coefficient of $\sigma_2[g]$ in local coordinates is not a differential polynomial of the type you want; rather, it is the square root of such a polynomial. Similarly, the trace of $C[g]^3$ is the cube of a $3$-form $\sigma_3[g]$, but, again, the coefficient of $\sigma_3[g]$ in local coordinates is not polynomial of the type that you want.
Finally, it can be shown by tensor analysis that, even if you consider arbitrary higher order derivatives, you cannot get a nontrivial polynomial scalar conformal invariant of the weight you want.
N.B. (Added note for clarity.) Even if one enlarges the possibilities by throwing in the square root of $\det(g_{ij})$ (which is useful in the Riemannian case to, say, define the volume form and the Hodge star, etc.) one still cannot get a nontrivial conformally invariant $3$-form $\sigma[g]$ such that, in local coordinates $x=(x^i)$ on $M^3$, the 3-form $\sigma[g]$ is of the form
$$\sigma[g] = F(g_{ij},\det(g_{ij})^{-1/\ell},g_{ij,k},g_{ij,kl},\ldots)
\,\mathrm{d}x^1\wedge \mathrm{d}x^2\wedge \mathrm{d}x^3,
$$
where $\ell>0$ is an integer and $F$ is some 'universal' polynomial in its arguments (finite in number).
The argument that this is so follows from general facts about the normal Cartan connection $\omega: B\to {\mathfrak{so}}(4,1)$, where $B\to M$ is the canonical parabolic bundle over $M$ associated to the conformal structure $[g]$, as defined by Cartan. (A reasonable reference for this might be Cap and Slovak's 2009 book Parabolic Geometries: Background and general theory.)
The point is that any differential invariant scalar $3$-form $\sigma[g]$ on $M$ would pull up to $B$ to be expressed as a semi-basic $3$-form in the canonical coframing whose coefficient $C_\sigma$ would be expressed as a 'universal' function of the $\omega$-curvature functions $C_{ijk}=-C_{ikj}$ on $B$ and their $\omega$-derivatives up to some finite order. If, in local coordinates, $\sigma[g]$ were polynomial of the above type, then the coefficient $C_\sigma$ would necessarily be a polynomial in the $C_{ijk}$ and their $\omega$-derivatives of weight $-3$. Now, the $C_{ijk}$ already have weight $-3$, and their $\omega$-derivatives have strictly lower weight, so the only possibility for a polyomial of weight $-3$ would be a linear expression in the $C_{ijk}$ alone. However, up to a constant multiple, the only $\mathrm{SO}(3)$-invariant linear expression is $C_{123}+C_{231}+C_{312}$, which vanishes identically.
