I don't think you need the analytic form of the folded Student's t-distribution, similarly to how you don't need the analytic form of the folded normal distribution to calculate the expected absolute value. If \(X\sim f\) and \(g\) is some function (which is sufficiently nice, etc.) then the expectation of \(g(X)\) is given by the law of the unconscious statistician:
\(\mathbf E[g(X)]=\int\limits_{-\infty}^\infty g(x)f(x)\,\mathrm dx\)
It is not too difficult to calculate the pdf of the folded distribution though: the distribution is symmetric, so the pdf of the folded distribution is just twice the pdf of the original one for nonnegative arguments and \(0\) for negative arguments. More generally the pdf of a folded distribution will be \(f+f\circ P\) for positive arguments and \(0\) for negative arguments, with \(f\) the original pdf and \(P\) parity change (multiplication by \(-1\)), i.e.
\(f_\text{folded}(x)=\begin{cases}0\quad &\text{for}\quad& x<0\\f(x)+f(-x) \quad&\text{for}\quad & x\geqslant 0\end{cases}\)
You may see this intuitively or with a proof: Let \(X\sim f\) with \(f\) a continuous pdf and cdf \(F\). Then we have that the cdf of \(|X|\) is given by
\(F_\text{folded}(x)=\mathbf P[|X|\leqslant x]=\mathbf P[-x\leqslant X\leqslant x]=\begin{cases}0\quad&\text{for}\quad&x<0\\F(x)-F(-x)\quad&\text{for}\quad& x\geqslant 0 \end{cases}\)
Differentiating we get that the pdf of \(|X|\), the folded pdf, is given by
\(f_\text{folded}(x)=\begin{cases}0\quad &\text{for}\quad&x<0\\F'(x)--F'(-x)=f(x)+f(-x)\quad &\text{for}\quad&x\geqslant 0\end{cases}\)
If \(z_{t-j}\) is distributed according to the standardised Student's t-distribution and \(f_\nu\) denotes the pdf of the standardised Student's t-distribution with \(\nu\) degrees of freedom, the expectation you are looking for is simply given by
