The following answer is for finite groups.
In characteristic zero, the group algebra is semisimple, so there are finitely many simple representations. These representations correspond to the blocks in the decomposition as a product of matrix algebras.
Here is a way to determine the number:
Start with the field $K$, adjoin the $g$th roots of unity ($g = |G|$) to get $L$, and consider the Galois group $\Gamma_K=L/K$. This is a subgroup of the multiplicative group of the integers mod $g$. Then $\sigma_t \in \Gamma_K$ corresponding to $t \in (\mathbb{Z}/g\mathbb{Z})^*$ acts on $G$ by raising $x \in G$ to the $t$-th power. The dimension of the space of class functions constant on $\Gamma_K$-orbits is the number of simple $K$-representations.
As for characteristic p, this is modular representation theory, The number of irreducibles is the number of $p$-regular conjugacy classes (where $p$-regular means the period is prime to $p$), when the field contains the $g$th roots of unity for $g$ the order of the group. See, e.g., Serre's Linear Representations of FInite Groups. My guess is that it should be true even without the assumption on the field being sufficiently large.
To answer your second question, there are soluble octics with the same Galois group involving
other $p$th roots of unity. Take $p\equiv 1\mod 7$, and $K={\mathbb Q}(\alpha)$
the unique degree 7 extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta_p)$. E.g. take
$\alpha=\sum_i \zeta_p^i$ where $i$ ranges over all seventh powers in ${\mathbb F_p}$.
If $f(x)$ is the minimal polynomial of $\alpha$ (degree 7), then $f(x^2)$ is the minimal
polynomial of $\sqrt\alpha$, which defines, generally, a 'random' quadratic extension of $K$.
That is, its Galois group over ${\mathbb Q}$ is
$$
G=C_2\wr C_7\cong C_2^7:C_7.
$$
Viewing $C_2^7$ as a 7-dimensional representation
of $C_7$ over ${\mathbb F}_2$, it decomposes as a 1-dimensional (trivial) representation plus
two distinct 3-dimensional ones. (The reason for this is that $2^3\equiv 1\mod 7$.) Factor out
$C_2^4\triangleleft G$, which is one of those
plus the trivial one. This gives a
Galois group $C_2^3:C_7$ that you are after,
and a subgroup $C_7$ in it cuts out the required octic field.
Here is a Magma code that can be used in the
Magma online calculator
to get such an octic:
p:=43; // or some other p = 1 mod 7
K<z>:=CyclotomicField(p);
alpha:=&+[z^i: i in [1..p] | IsPower(GF(p)!i,7)];
R<x>:=PolynomialRing(Rationals());
f:=Evaluate(MinimalPolynomial(alpha),x^2);
K:=NumberField(f);
assert exists(a){a: a in ArtinRepresentations(K) | #Kernel(Character(a)) eq 16};
F:=Field(Minimize(a));
DefiningPolynomial(F);
You can also stick in a Tschirnhaus transformation, say,
alpha:=alpha^3+alpha+1;
in the 5th line to vary the generator of $K$ - in this way you get all possible
$C_2^3:C_7$-extensions involving $p$th roots of unity.
For your questions in the comments, the roots might be real or complex, and the constant term
may or may not be a square - this depends on whether $\alpha$ is chosen to be totally real,
and on the way 'Minimize' works; you can always use an additional Tschirnhaus transformation
to modify the final output or Pari's 'polredabs' function to try and get the coefficients smaller.
I do not know the reason why for $p=29$ there is such an elegant octic, this is very curious. It is a bit like the Trinks polynomial $x^7-7x+3$ with Galois group PSL(2,7), and I wonder whether simple polynomials having interesting Galois group is such a statistical blip, or there is a reason behind it.
Best Answer
If such polynomials exist, there will only be finitely many of them, up to composing on both sides with scalar polynomials $\alpha x$ with $\alpha\in\mathbf{Q}$. More generally, Guralnick and Shareshian proved that if $d=7$ or $d>8$ then there are only finitely many equivalence classes of irreducible degree-$d$ trinomials in $\mathbf{Q}[x]$ whose Galois group is not $A_d$ or $S_d$, where two trinomials $f(x)$ and $g(x)$ are called equivalent if $f(x)=\alpha\cdot g(\beta x)$ for some $\alpha,\beta\in\overline{\mathbf{Q}}$. They proved an analogous result over number fields, and also for reducible trinomials. See Theorem 1.4.3 of "Symmetric and Alternating Groups as Monodromy Groups of Riemann Surfaces I: Generic Covers and Covers with Many Branch Points", AMS Memoirs (2007), vol. 189, no. 886.
This explains why you were able to find the examples you found. The only pairs $(d,n)$ for which there are infinitely many equivalence classes of irreducible degree-$d$ trinomials in $K[x]$ with terms of degrees $0,n,d$ and Galois group not $A_d$ or $S_d$ (for some number field $K$) are those pairs with $0<n<d$ and $d\le 8$ and $d\ne 7$ and either $d=4$ or $\gcd(n,d)=1$. Also Guralnick-Shareshian listed the groups which occur for infinitely many equivalence classes over a number field.
In order to answer the question you asked, one can use the following approach by Elkies. Let $H=\text{AGL}(1,7)$ be the group of all linear maps on $\mathbf{F}_7$. For each $n\in\{1,2,\dots,6\}$, Guralnick-Shareshian say that Elkies told them that there is a bijection between the set of equivalence classes of solvable degree-$7$ trinomials in $\mathbf{Q}[x]$ with terms of degrees $0,n,7$ and the set of rational points on a certain Riemann surface $X$. The information Guralnick-Shareshian use about this Riemann surface is that it admits a degree-$120$ branched cover $X\to\mathbf{P}^1$ with monodromy group $S_7$ and three branch points, two of which have branch cycles $(1,n+1)$ and $(1,2,\dots,7)$. I haven't thought about how to prove this bijection. Anyway, assuming it's correct, then your question amounts to asking whether any of six specific Riemann surfaces have any rational points.
Added later: as Noam notes in his comment, there are really three Riemann surfaces, because the ones for $n$ and $7-n$ are the same. By my computations, all of these Riemann surfaces $X$ are defined over $\mathbf{Q}$, and the genus of $X$ is $17$, $10$, or $16$ according as $n=1$, $2$, or $3$. So the original question amounts to asking whether there are rational points on any of three specific curves over $\mathbf{Q}$, having genera $17$, $10$ and $16$.