Let me give you a few basic facts about of representation theory of $SO(N)$ groups (set $N=d-1$ for the situation at hand; also, we should be talking about the universal cover $Spin(N)$, but to be more general let's pretend that we are working with complex Lie algebra $so(N,\mathbb{C})$ -- the representation theory over $\mathbb{C}$, in which one is interested in this case, is independent of signature, and is equivalent to that of $so(N,\mathbb{C})$). There are two different cases, $N=2n$ and $N=2n+1$. In the former case the Lie algebra of $SO(N)$ is of $D_n$ type and in the latter of $B_n$ type; the representation theory is slightly different in these two cases.
The first thing to know is that $SO(N)$ is reductive for any $N$ -- that is, all its finite dimensional representations decompose as a direct sum of irreducible representations, so you only need to know what are the irreducible representations. The second thing to know is that for $N>3$ ($d>4$) there is no single "spin" which labels the representations of $SO(N)$, but instead you need to specify the weight, which is a vector of $n$ numbers (recall $N=2n$ or $N=2n+1$). That is just how the life works, and there is no way around this. In particular, for $N>3$ spinor and vector representations are not naturally members of the same family labeled by a single spin, so there is no reason to expect that dimension of spinor representation will be smaller than the dimension of vector representation (and this is in fact false).
These $n$ numbers can be taken to be the so-called Dynkin labels, and we can write them as a list $[\lambda_1,\ldots,\lambda_n]$. The $\lambda_i$ are all independent non-negative integers. To show you the basic idea, let us consider $SO(5)$ and $SO(6)$ with $n=2$ and $n=3$ respectively. For $SO(5)$ the simplest representations, which one usually calls spin-$j$, are $[j,0]$. These are rank-$j$ traceless totally symmetric tensors. The spinor representation is, however, $[0,1]$, so it is not a part of this family. For $SO(6)$ we also have the family $[j,0,0]$ and two chiral spinor representations $[0,1,0]$ (left-handed) and $[0,0,1]$ (right-handed). As you can see, spinors are again not a part of the family. More generally, for $N=2n+1$ the last label is the "spinor label", and for $N=2n$ the last two labels are the "spinor labels".
Now, let me stress that you cannot set $j=\frac{1}{2}$ above, because the labels must be integers. You can in $SO(3)$ because there we have the vector $[\lambda_1]$ and $\lambda_1$ is the spinor label and the traceless-symmetric "spin-$j$" family is $[2j]$, a low-dimensional peculiarity. (There is also a similar
peculiarity in $SO(4)$ where the traceless-symmetric irreps are $[j,j]$.) Furthermore, there are other representations besides those mentioned above (you can set all $\lambda_i$ to whatever you want), but they have more complicated interpretations ($p$-forms, mixed-symmetry tensors, spin-tensors, etc).
Important thing to know is, however, that the Fermi-Bose statistics is determined by the parity of $\lambda_n$ ($N=2n+1$) or $\lambda_n+\lambda_{n-1}$ ($N=2n$). Thus to get a fermionic representation it is crucial to set at least one of spinor labels to a non-zero value.
Finally, let me answer directly your question about the dimensions of representations. Hopefully, it is clear by now that you cannot get away with just one number $s$, and instead you need all the $n$ labels $\lambda_i$. Correspondingly, the answer is quite complicated. (Consult any representation theory text for the formula.) The simple special cases are:
- The irreducible spinor representations are $2^n$ for $N=2n+1$ and $2^{n-1}$ for $N=2n$. In the latter case I am talking about Weyl spinors and a Dirac spinor is a sum of left and right Weyl spinors, hence not irreducible.
- Traceless-symmetric spin-$j$ representation has the dimension $$
\dim [j,0,\ldots,0] = \frac{\Gamma(j+N-2)(2j+N-2)}{\Gamma(j+1)\Gamma(N-1)} = \frac{(j+N-3)!(2j+N-2)}{j!(N-2)!}.
$$
