I would suggest that this is a problem with how the results are reported. Not to "beat the Bayesian drum" but approaching model uncertainty from a Bayesian perspective as an inference problem would greatly help here. And it doesn't have to be a big change either. If the report simply contained the probability that the model is true this would be very helpful. This is an easy quantity to approximate using BIC. Call the BIC for the mth model $BIC_{m}$. Then the probability that mth model is the "true" model, given that $M$ models were fit (and that one of the models is true) is given by:
$$P(\text{model m is true}|\text{one of the M models is true})\approx\frac{w_{m}\exp\left(-\frac{1}{2}BIC_{m}\right)}{\sum_{j=1}^{M}w_{j}\exp\left(-\frac{1}{2}BIC_{j}\right)}$$
$$=\frac{1}{1+\sum_{j\neq m}^{M}\frac{w_{j}}{w_{m}}\exp\left(-\frac{1}{2}(BIC_{j}-BIC_{m})\right)}$$
Where $w_{j}$ is proportional to the prior probability for the jth model. Note that this includes a "penalty" for trying to many models - and the penalty depends on how well the other models fit the data. Usually you will set $w_{j}=1$, however, you may have some "theoretical" models within your class that you would expect to be better prior to seeing any data.
Now if somebody else doesn't report all the BIC's from all the models, then I would attempt to infer the above quantity from what you have been given. Suppose you are given the BIC from the model - note that BIC is calculable from the mean square error of the regression model, so you can always get BIC for the reported model. Now if we take the basic premise that the final model was chosen from the smallest BIC then we have $BIC_{final}<BIC_{j}$. Now, suppose you were told that "forward" or "forward stepwise" model selection was used, starting from the intercept using $p$ potential variables. If the final model is of dimension $d$, then the procedure must have tried at least
$$M\geq 1+p+(p-1)+\dots+(p-d+1)=1+\frac{p(p-1)-(p-d)(p-d-1)}{2}$$
different models (exact for forward selection), If the backwards selection was used, then we know at least
$$M\geq 1+p+(p-1)+\dots+(d+1)=1+\frac{p(p-1)-d(d-1)}{2}$$
Models were tried (the +1 comes from the null model or the full model). Now we could try an be more specific, but these are "minimal" parameters which a standard model selection must satisfy. We could specify a probability model for the number of models tried $M$ and the sizes of the $BIC_{j}$ - but simply plugging in some values may be useful here anyway. For example suppose that all the BICs were $\lambda$ bigger than the one of the model chosen so that $BIC_{m}=BIC_{j}-\lambda$, then the probability becomes:
$$\frac{1}{1+(M-1)\exp\left(-\frac{\lambda}{2}\right)}$$
So what this means is that unless $\lambda$ is large or $M$ is small, the probability will be small also. From an "over-fitting" perspective, this would occur when the BIC for the bigger model is not much bigger than the BIC for the smaller model - a non-neglible term appears in the denominator. Plugging in the backward selection formula for $M$ we get:
$$\frac{1}{1+\frac{p(p-1)-d(d-1)}{2}\exp\left(-\frac{\lambda}{2}\right)}$$
Now suppose we invert the problem. say $p=50$ and the backward selection gave $d=20$ variables, what would $\lambda$ have to be to make the probability of the model greater than some value $P_{0}$? we have
$$\lambda > -2 log\left(\frac{2(1-P_{0})}{P_{0}[p(p-1)-d(d-1)]}\right)$$
Setting $P_{0}=0.9$ we get $\lambda > 18.28$ - so BIC of the winning model has to win by a lot for the model to be certain.