Expected value of the log of a sum of half normal distributions

given $X_1, \ldots, X_n$ independent random variables following a centered gaussian distribution $\mathcal{N}(0, 1)$, I wish to find the expected value of $Y = \log(|X_1|+\ldots + |X_n|)$ (having the expression of the pdf would also be great).

Using the integral expression, \begin{align}\mathbb{E}(Y) &= \int_{x=-\infty}^{+\infty} \int_{x_1=0}^{e^x} \cdots \int_{x_{n-1}=0}^{e^x - \sum_{i=1}^{n-2} x_i} x \sqrt{\frac{2}{\pi}} e^{-x_1^2/2} \cdot \cdots \cdot \sqrt{\frac{2}{\pi}} e^{-x_{n-1}^2/2} \cdot \sqrt{\frac{2}{\pi}} e^{-(e^x - \sum_{i=1}^{n-1} x_i)^2/2} dx_{n-1} \ldots dx \\& = \left(\sqrt{\frac{2}{\pi}}\right)^n \int_{t=0}^{+\infty} \frac{\log t}{t} e^{-t^2/2}\int_{x_1 = 0}^{t} \cdots \int_{x_{n-1}=0}^{t - \sum_{i=1}^{n-2} x_i} e^{t(\sum_{i=1}^{n-1} x_i) - \sum_i x_i^2} dx_{n-1} \ldots dt \\& = \left(\sqrt{\frac{2}{\pi}}\right)^n \int_{t=0}^{+\infty} \frac{\log t}{t} e^{-t^2/4}\int_{x_1 = 0}^{t} \cdots \int_{x_{n-2}=0}^{t - \sum_{i=1}^{n-3} x_i} e^{t(\sum_{i=1}^{n-2} x_i) - \sum_{i=1}^{n-2} x_i^2} \frac{\sqrt{\pi}}{2}\left(\text{erf}\left(\frac{t}{2} - \sum_{i=1}^{n-2} x_i\right) + \text{erf}\left(\frac{t}{2}\right)\right) dx_{n-1} \ldots dt\end{align}but I am a bit unsure how to proceed from there; is there a simpler way to go about this problem? Thank you in advance!

I have also noted there is a way to simplify the integral with $t = \sum_i x_i$, which provides:\begin{align}\mathbb{E}(Y) & = \left(\sqrt{\frac{2}{\pi}}\right)^n \int_{t=0}^{+\infty} \frac{\log t}{t} e^{t^2/2}\int_{x_1 = 0}^{t} \cdots \int_{x_{n-1}=0}^{t - \sum_{i=1}^{n-2} x_i} e^{-\sum_i x_i^2} dx_{n-1} \ldots dt,\end{align}and I think I should be able to sort the computation out with a change of variables from the sphere for the euclidian norm and the one for norm-1...