First note that the vector $x$ distributed uniformly on the $d$-dimensional hypersphere can be constructed from a vector $y$ with i.i.d. normal elements $y_1,y_2,\ldots y_d$, via
$$x=\left(\sum_{i=1}^d y_i^2\right)^{-1/2} y.$$
Consider a set of $n$ vectors $y^{(1)},y^{(2)}\ldots y^{(n)}$ and construct the $n\times d$ dimensional matrix $Y$ with elements $Y_{ij}=y^{(i)}_j$. The matrix product $Y^TY$ has a Wishart distribution, and $(Y^TY)^+$ has the inverse Wishart distribution, with expectation value
$$\mathbb{E}(Y^TY)^+=\frac{I}{n-d-1},\;\;n>d+1.$$

Also construct the $n\times n$ diagonal matrix $D$ with elements $D_{ij}=\delta_{ij}\left(\sum_{k=1}^d (y_k^{(i)})^2\right)^{-1/2}$. The matrix $X$ in the OP is related to the matrix $Y$ by $X=DY$.

We seek the expectation value of the scalar
$$R\equiv \frac{1}{d}\,{\rm tr}\,(X^TX)^+=\frac{1}{d}{\rm tr}\,(Y^T D^2Y)^+.$$
For $d\gg 1$ the matrix $D^2$ selfaverages to $1/d$ times the identity, hence we estimate
$$r_{n,d}\equiv\mathbb{E}(R)\approx\frac{d}{n-d-1},\;\;n>d+1\gg 1.$$

I have performed some numerical checks, averaging over 500 realizations.

d,n |
d/(n-d-1) |
numerics |

100,105 |
25 |
25.1 |

100,110 |
11.11 |
11.0 |

100,120 |
5.26 |
5.15 |

50,100 |
1.02 |
1.00 |

50,200 |
0.336 |
0.332 |

The difference between the numerical value and the estimate $d/(n-d-1)$ is within the statistical uncertainty.

whichscalar... (For wishart this is known) $\endgroup$1more comment