Here is yet another overkill, but hopefully not too bad a way to prove this inequality.
We have the following proof sketch.
$$\begin{eqnarray} x^T(A+B)x &\ge& x^TAx\quad\forall x\\\\ -x^T(A+B)x &\le& -x^TAx\\\\ \exp(-x^T(A+B)x) &\le& \exp(-x^TAx)\\\\ \int\exp(-x^T(A+B)x)dx &\le& \int\exp(-x^TAx)dx\\\\ \frac{1}{\sqrt{\det(A+B)}} &\le& \frac{1}{\sqrt{\det(A)}}\\\\ \det(A+B) &\ge& \det(A) \end{eqnarray} $$
The only fancy thing that happened is in the second last line, where I used the formula for the [Gaussian integral (see multivariate section)][1]
Update. To expand upon my comment below, to note that the above idea actually with a little bit more care actually yields a proof of the Minkowski determinant inequality, by equivalently establishing log-concavity of the determinant. The key point to observe is \begin{eqnarray} \exp(-x^T((1-\lambda)A+\lambda)x) &=& [\exp(-x^TAx)]^{1-\lambda}[\exp(-x^TBx)]^\lambda\\\\ \int\exp(-x^T((1-\lambda)A+\lambda)x)dx &=& \int [\exp(-x^TAx)]^{1-\lambda}[\exp(-x^TBx)]^\lambda\ dx\\\\ &\stackrel{\text{Hölder}}{\le}& \left(\int\exp(-x^TAx)dx \right)^{1-\lambda}\left(\int \exp(-x^TBx)dx \right)^\lambda. \end{eqnarray} Now invoke the Gaussian integral as above to conclude \begin{equation*} \det((1-\lambda)A+\lambda B) \ge \det(A)^{1-\lambda}\det(B)^\lambda, \end{equation*} from which we can easily conclude $\det(A+B)^{1/n} \ge \det(A)^{1/n}+\det(B)^{1/n}$. [1]: http://en.wikipedia.org/wiki/Gaussian_integral