Let $\def\B{\mathrm{B}}\B(t)$ be usual Brownian motion, and let $\B_b(t)$ be the Brownian bridge. Explcitly, $$\B_b(t)=\B(t)-tB(t).$$
It can be shown (see A Relation between Brownian Bridge and Brownian Excursion) that
$$
\B_e(t)\stackrel{D}= B_b(\tau_m+t\!\!\mod 1)-B_b(\tau_m),
$$
where $\tau_m$ is the time when $\B_b$ attains its minimum. It follows that
$$
E\left[\int_0^1\B_e(t)\,dt\right]=E\left[\int_0^1\B_b(t)\,dt\right]-E\left[\B_b(\tau_m)]=0-E[\B_b(\tau_m)\right],
$$
Therefore, the expectation you want is the opposite of the expected minimum of a Brownian bridge, or equivalently the expected maximum of a Brownian bridge.
Let $M=\sup_{0\le t\le 1}\B_b(t)$ be this maximum. Using the reflection principle, you can show that
$$
P(M> m)=P(\B_1=2m\mid \B_1=0)=\frac{
\frac1{\sqrt{2\pi}}\exp(-(2m)^2/2)}{\frac1{\sqrt{2\pi}}\exp(-0^2/2)}=\exp(-2m^2).
$$
Namely, if you take a Brownian bridge and reflect it after the time it reaches $+m$, you get a Brownian motion path which ends at $2m$. Conversely, any path with reaches $2m$ reflects to a bridge which attains $m$. Of course, conditioning on a zero probability event is not allowed, but the above is the intuition which can be made rigorous.
Putting this altogether,
$$
E\left[\int_0^1\B_e(t)\,dt\right]=E[M]=\int_0^\infty P(M>m)=\int_0^\infty \exp(-2m^2)\,dm=\sqrt{\pi/8}.
$$
