A compact flat manifold whose first Betti number is equal to the dimension is a flat torus I know the following to be true:
If $(M,g)$ is a compact flat Riemannian manifold whose first Betti number ($= \dim H_{dR}^1(M)$) is equal to
the dimension, then it is (isometric to) a flat torus.
In the notes I am following it is stated that this follows from Bieberbach's Theorem (I guess it is the one about flat manifolds being finitely covered by tori), yet I have not found the way to link it to the first Betti number.
Can someone show me how to prove the statement using Bieberbach's Theorem?
 A: I'm not sure of the theorem you quote but here is my understanding (apologies to Mods if this is seen as not an answer to the OP).
Let $(M^n, g)$ be Riemannian manifold and $T^2$ a torus. Let $b_1$ be the first Betti number of our manifold $M$. Using an integral basis of
harmonic 1-forms we can define, by integration, the Jacobi map $\pi$; this in turn gives a map onto a torus $T^{b_1}$, on which we put the usual flat metric. 
In our case, as all harmonic $1$-forms are of constant length, we are allowed to choose this basis to be pointwise orthonormal. Therefore $\pi$ is a Riemannian submersion and besides we must have $b_1 \leq n$.
Proposition 
Let $(M^n, g)$ be a compact Riemannian manifold and $b_1$ be its first
Betti number. Then all harmonic $1$-forms are of constant length if and only if
$(M^n, g)$ is a locally trivial fibre bundle, with minimal fibres, over a $b_1$-dimensional flat torus and such that $b_1 \leq n$. Moreover under the sequence
$$F^{n-b_1} \hookrightarrow M^n \xrightarrow{\pi}T^{b_1}$$
we see that if $b_1 = n$, then $\pi$ is a Riemannian isometry and hence $(M^n, g)$ is a flat torus.
