Case $d=3$:

The result becomes inconsistent unless we make $\lambda\to 0$ in the continuum limit, which takes us to the Gaussian point. If we do that sufficiently fast then we get $J_{0}=m_{\mathfrak{N}}^{2}V_{0}$, a result consistent with a linear theory. Otherwise, if we take $\lambda$ to zero at the appropriate rate, we get

$$J_{0} = V_{0} \left( m_{\mathfrak{N}}^{2} - 2\Lambda\, V_{0}^{2} \right).$$

Since the sign of the second term is reversed, this result does not seem to make much sense, even if we consider that the $\Lambda$ that appears there is a bare parameter, a parameter characterizing a continuum-limit flow, in fact, and not the renormalized coupling constant.

Observe however that this is not necessarily a free theory, even at the Gaussian point, where one would expect that $\lambda_{R}=0$. This is so because in $d=3$ we have $\Lambda_{R}=\lambda_{R}/a$, and therefore we may have $\Lambda_{R}\neq 0$ even if $\lambda_{R}\to 0$ as we take the limit and make $a\to 0$. In other words, in $d=3$ we may have interacting limits of this type sitting right on top of the Gaussian point.

The complete failure of the approximation away from the Gaussian point suggests that in that case the distribution of the $d=3$ model may not be sufficiently close to a Gaussian distribution to allow a Gaussian approximation to work, and hence its expectations values cannot be well represented by the Gaussian measure of $S_{0}[\vec{\varphi}']$.