Fleshing out Wojowu's comment: set $K = \big\lfloor \frac{\log n}{\log p} \big\rfloor$. Since
\begin{align*}
v_p(n!) = \sum_{1\le k\le K} \bigg\lfloor \frac n{p^k} \bigg\rfloor &\ge \sum_{1\le k\le K} \bigg( \frac n{p^k} - 1 \bigg) \\
&= \frac n{p-1} - \frac n{p^K(p-1)} - K = \frac n{p-1} + O\bigg( \frac{\log n}{\log p} \bigg)
\end{align*}
for $p\le n$, we have
\begin{align*}
\log\prod_{p\equiv 3 \pmod 4} p^{v_p(n!)} &= \sum_{\substack{p\le n \\ p\equiv 3\pmod 4}} \bigg( \frac n{p-1} + O\bigg( \frac{\log n}{\log p} \bigg) \bigg) \log p \\
&= n \sum_{\substack{p\le n \\ p\equiv 3\pmod 4}} \frac{\log p}{p-1} + O\big( \pi(n;4,3) \log n \big) \\
&= \frac{n \log n}2 + O(n\log n),
\end{align*}
where the last equality used partial summation.

In hindsight, of course $n(\log n)/2$ should be the main term: we expect the product to be roughly the square root of $n!$, and $\log\sqrt{n!} \sim n(\log n)/2$ by Stirling's formula.

All the steps of this argument can be given with explicit constants in the inequalities if you want (including corresponding upper bounds); the partial summation step can start with an explicit lower bound for $\theta(n;4,3)$ found in [this paper](http://www.math.ubc.ca/~gerg/papers/downloads/EBPAP.pdf) for example.