As Kevin Walker pointed out in a comment, Dijkgraaf-Witten invariants are weighted by $1/\mathrm{Stab}(\rho)$. In the same way, the Yetter invariant for $\mathcal T$ and $M$ is generally defined such that it's weighted using the 2-groupoid cardinality of $\pi_{\le 2}\mathrm{Map}(M, \mathcal T)$, so that the invariant is $$\sum_{[f\colon M\to\mathcal T]} \frac{\pi_2(\mathrm{Map}(M, \mathcal T), f)}{\pi_1(\mathrm{Map}(M, \mathcal T), f)}.$$$$\sum_{[f\colon M\to\mathcal T]} \frac{|\pi_2(\mathrm{Map}(M, \mathcal T), f)|}{|\pi_1(\mathrm{Map}(M, \mathcal T), f)|}.$$
If we use this normalization, the Yetter invariants are the partition functions of a TQFT $Z_{\mathcal T}$, usually called the Yetter model. In this case, a different MathOverflow answer by Kevin Walker tells us that $Z_{\mathcal T}$ is multiplicative under connect sum iff
- $\dim Z_{\mathcal T}(S^{n-1}) = 1$, and
- $Z_{\mathcal T}(S^n) = 1$.
The state space $Z_{\mathcal T}(M^{n-1}) := \mathbb C[[M, \mathcal T]]$, so for $n = 3$, the first property doesn't hold: $\dim Z_{\mathcal T}(S^2) = |\pi_2(\mathcal T)|$. A similar problem occurs for $n = 2$.
If $n > 3$, then $[S^{n-1}, \mathcal T] = 0$, so the first property holds. The second property does not quite hold: $[S^n, \mathcal T] = 0$, but we have to calculate the weighting. Since $\mathrm{Map}(S^n, \mathcal T)\simeq\mathrm{Map}(\mathrm{pt}, \mathcal T)\cong \mathcal T$, $$Z_{\mathcal T}(S^n) = \frac{\pi_2(\mathcal T)}{\pi_1(\mathcal T)},$$$$Z_{\mathcal T}(S^n) = \frac{|\pi_2(\mathcal T)|}{|\pi_1(\mathcal T)|},$$ which is frequently not equal to 1.
I don't know about the unweighted version you mentioned, since it doesn't come from a TQFT as far as I know.