Why do the even Bernoulli numbers grow so fast? Question is in the title.
We have:
$$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$

Recall the generating function
$$f(z) = \sum_{q\ge 0} B_q \frac{z^q}{q!} = \frac{z}{e^z-1}.$$
We can treat it as if it were rational as the poles are countable with
no accumulation point, so we may use the techniques of extracting coefficients from rational generating functions, with the asymptotics being given by inverse powers of the dominant poles.
The pole at $z=0$ is cancelled by the factor $z$ in the numerator. The
nearest two poles are at $\pm 2\pi i,$ both at the same distance from zero ($2\pi$).
Computing the residues we obtain
$$\mathrm{Res}(f(z); z=\pm 2\pi i) = \pm 2\pi i$$
and from this we get the singular decomposition
$$\frac{2\pi i}{z-2\pi i} - \frac{2\pi i}{z+2\pi i}.$$
Turn this into geometric series to get
$$\frac{1}{z/(2\pi i)-1} - \frac{1}{z/(2\pi i)+1}
= - \frac{1}{1-z/(2\pi i)} - \frac{1}{1+z/(2\pi i)}
\\= -\sum_{q\ge 0} \frac{z^q}{(2\pi i)^q}
-\sum_{q\ge 0} (-1)^q \frac{z^q}{(2\pi i)^q}
= -2 \sum_{q\ge 0} \frac{z^{2q}}{(2\pi i)^{2q}}
\\ = -2 \sum_{q\ge 0} (-1)^q \frac{z^{2q}}{(2\pi)^{2q}}
= 2 \sum_{q\ge 0} (-1)^{q+1} \frac{z^{2q}}{(2\pi)^{2q}}.$$
Extracting coeffcients and since $B_{2n}  = (2n)! [z^{2n}] f(z)$ we find that
$$B_{2n} \sim (2n)! \times 2 \times 
\frac{(-1)^{n+1}}{(2\pi)^{2n}}.$$
Using Stirling's formula this becomes
$$\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}
\times 2 \times
\frac{(-1)^{n+1}}{(2\pi)^{2n}}.$$
This is
$$4 \times (-1)^{n+1} \times \sqrt{\pi n} 
\left(\frac{n}{\pi e}\right)^{2n}.$$
Addendum. To  justify rigorously the above asymptotics  we need to
verify that  there isn't an additional entire  component not accounted
for in the complete singular decomposition.
To do this introduce the sum
$$g(w) = 
\sum_{q\ge 1}\left(\frac{1}{w/q-1}-\frac{1}{w/q+1}\right)
= 2 \sum_{q\ge 1} \frac{1}{w^2/q^2-1}.$$
This  can  be  evaluated  using  a  standard  technique  from  complex
variables and setting
$$G(z) = 1/(w^2/z^2-1)\times\pi\cot(\pi z)$$
we have
$$g(w) = 
-
\left(
\mathrm{Res}(G(z); z=0)
+\mathrm{Res}(G(z); z=w)
+\mathrm{Res}(G(z); z=-w)
\right).$$
Computing the residues we get
$$g(w) = \pi w \cot(\pi w).$$
Now we claim that
$$f(z) = -\frac{1}{2}z + \sum_{q\ge 1} 
\left(\frac{2\pi i q}{z-2\pi i q}-\frac{2\pi i q}{z+2\pi i q}\right).$$
The sum is $g(z/(2\pi i))$ so we get
$$-\frac{1}{2}z + \frac{z}{2i}  \cot\left(\frac{z}{2i}\right)
= -\frac{1}{2}z + \frac{z}{2i}  \times i \times 
\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}
\\= \frac{1}{2} z
\left(-1 + \frac{e^z+1}{e^z-1}\right)
= \frac{1}{2}z \frac{2}{e^z-1} = \frac{z}{e^z-1}.$$
This shows  rigorously that the  entire part is  $$-\frac{1}{2}z$$ and
hence makes no  contribution to $B_{2n}$ where $n\ge  1$ and the above
singular decomposition is justified.
