Actually it is quicker to sketch the proof than checking a reference.
Assume $u:\mathbb{R}^n\to \mathbb{R}$ is measurable and periodic w.r.to $x_i$ with period $b_i - a_i$, for $1\le i\le n$. Then by a linear change of variables $\|u\| _ {p,M}=\|u _ \nu \| _ {p, M} $ for $1\le p\le\infty$. So, if $u\in L^p_{loc}$ the sequence $\{u _ \nu \} _ {\nu\in\mathbb{N} } $ is bounded in $L^p(M)$. As a general fact, when checking the weak (or weak*) convergence of a bounded sequence in a Banach space $E$ (respectively, in its dual), a norm-dense set of test elements of $E^*$ (resp. $E$) is sufficient. Here, the thesis easily follows using as test functions the
continuous functions on $M:=\prod _ {i=1}^n [a _i, b _ i]$ , for which it holds
$$\left | \int_M u _ \nu(x) \phi(x)dx - \int_M \tilde u \phi(x)dx \right | \leq \|u\|_{1,M}\, \omega( {\delta}/{\nu})\, ,$$
where $\omega$ is a modulus of continuity for the uniformly continuous function $\phi$ on $M$ and $\delta$ is the diameter of $M$ (write $\int_M \tilde u \phi(x)dx$ as $\int_M u _ \nu(x) \hat \phi(x)dx$ with a suitable discretization $\hat \phi$ of $\phi$). Note that, of course, the argument wouldn't work for $p=1$, as continuous functions are not dense in $L^\infty$.
More details. Let $u\in L^1 _ {loc}$ with parallelepiped-period $M$, let $\phi \in C^0(M)$, and $\nu \in \mathbb{N} _ +$. First observe that $M$ is partitioned into $\nu^{n}$ smaller parallelepipeds, say $\{ M_j\} _ {1\le j \le \nu^n}$, which are translated copies of $\nu ^{-1} M$, and have in particular diameter $\delta / \nu$, where as above $\delta:=\mathrm{diam}(M)\\ $ . Define a simple function $\hat\phi$ that takes a constant value on each of these $M _ j$, namely $$\hat \phi _ {|M _ j}:= \frac{1}{|M _ j|}\int _ { M _ j}\phi (x)dx,\quad \mathrm{for\, any\quad } 1\le j \le \nu^n .$$ Note that by continuity of $\phi$, $\hat\phi _ {|M _ j}=\phi(\xi _ j)$ for some $\xi _ j\in M _ j $, whence $\| \hat \phi - \phi\| _ {\infty, M}\le \omega (\delta / \nu)$ . Moreover, the integral mean of the function $u _ \nu$ over each $M _ j$ is $\frac{1}{|M _ j|}\int _ {M _ j} u _ \nu (x) dx = \frac{1}{|\nu M _ j|} \int _ {\nu M _ j } u(x) dx=\tilde u $, because $\nu M _ j$ is a translated copy of the period $M$, so that the integral of $u$ over them is the same. Then, since $\hat\phi$ is constant on each $M _ j$ we have
$$ \int_M u _ \nu(x) \hat \phi(x)dx = \sum _ {j=1} ^{\nu\, ^n} \int _ {M _ j} u _ \nu(x) \hat \phi(x)dx= \sum _ {j=1} ^{\nu\, ^n}\, \bigg(\frac{1}{|M _ j|}\int _ { M _ j}\phi (x)dx\bigg) \int _ {M _ j} u _ \nu(x) dx=$$
$$= \sum _ {j=1} ^{\nu\, ^n}\, \tilde u\,, \int _ { M _ j}\phi (x)dx = \tilde u \int _ M \phi(x) dx\, .$$
Thus,
$$\left | \int _ M u _ \nu(x) \phi(x)dx - \int _ M \tilde u \phi(x)dx \right| = \left | \int _ M u _ \nu(x) \big( \phi(x) - \hat \phi(x)\big)dx \right| \le \|u\|_{1,M}\, \omega( {\delta}/{\nu})=o(1)\, ,$$
as $\nu\to\infty$. This is interpretated: if $u\in L^1 _ {loc}$, then $u _ \nu$ converges to $\tilde u$ in the weak* sense of measures on $M$. Moreover, as remarked above, it follows immediately that if $u\in L^p _ {loc}$ for some $1 < p < \infty$, the convergence is weak $L^{p}(M)$, and if $u\in L^\infty _ {loc}$ the convergence is weak* $L^\infty (M)$.
