Here are the first two cases (both with $d=2$). I mention them and the (fairly simple) method in the hope that some more subtle analysis will yield more results. I don't know if one must derive the distribution of region sizes but, if so, that is non-trivial in itself even when $d=2.$

For $d=2$ and $n=1,$ the unit circle is split in two equal halves. The probability that all $k$ points are in the same half (as the first one) is $\frac{1}{2^{k-1}}.$ So the expected number is $$2-\frac{1}{2^{k-1}}.$$

For $d=2$ and $n=2,$ I get an expected number of $$ 4-\frac{8}{k+1}+\frac{4}{2^k(k+1)}=4\left(1-\frac{1}{2^{k+1}}\right)\left(1-\frac{2}{k+1}\right)-\frac{1}{2^{k-1}} .$$

I'm not sure what the near factorization adds, but there it is.

Analysis: Note: I did this first ,but was reassured when a simulation with $100000$ trials and $k=6$ gave close to the predicted values. The circle is split into four pieces of (relative) sizes $\alpha,\frac{1}{2}-\alpha,\alpha,\frac{1}{2}-\alpha$ with $\alpha$ uniformly distributed in $[0,\frac{1}{4}]$
So the probability that $k$ points all land in the same piece is
$$p_1=\int_0^{1/4}\alpha^k+ (\frac{1}{2}-\alpha)^k+\alpha^k+(\frac{1}{2}-\alpha)^kd\alpha=\frac{4}{(k+1)2^k}.$$

There are six ways to pick two regions: four of combined size $\frac{1}{2}$ and one each of sizes $2\alpha$ and $1-2\alpha.$ Accordingly, the probability that all $k$ points land in exactly two regions is

$$p_2=\int_0^{1/4}(2\alpha)^k+ 4(\frac{1}{2})^k+(1-2\alpha)^kd\alpha-3p_1=\frac{2}{k+1}+\frac{4}{2^k}-\frac{12}{(k+1)2^k}.$$
The subtraction is to compensate for the fact that an event that all the points actually land in one region gets counted $3$ times by the integral as being in the union of two regions.
There are only two possible sizes for the combined length of $3$ regions leading to $$p_3=\int_0^{1/4}2(\frac{1}{2}+\alpha)^k+2(1-\alpha)^k d\alpha-2p_2-3p_1=\frac{4}{k+1}-\frac{8}{2^k}+\frac{8}{(k+1)2^k}$$ and $$p_4=1-p_1-p_2-p_3=1-\frac{6}{k+1}+\frac{4}{2^k}.$$

Then $p_1+2p_2+3p_3+4p_4$ gives the expected number.

For $k=6$ and $100,000$ trials this would predict frequencies of $$[893,32143,46429,20536]$$ (everything rounds up) and a simulation gave $$[924,32079,46445,20552].$$

For arbitrary $d,n$ the regions come in $R=\sum_{i=0}^{d-1} {n -1 \choose i}$ sizes with two of each size. A (highly?) nontrivial question is what the distribution function (over some $R-1$ simplex) is. For $d=2$, $R=n$ and for $d=3,$ $R=\frac{n^2-n+2}{2}.$ Perhaps for $n=3$ or $4$ iterated integrals as above could be pushed through. For $d=2$ and large $n$, one has a Poisson process.