Generalizing Pedro's comment somewhat, one can think about the relationship between the Frobenius theorem and curvature as follows.
Let $M$ be a smooth real $m$-manifold, and $H\subseteq TM$ a smooth $k$-dimensional ($0<k<m$) distribution on $M$. The Frobenius theorem then states that $H$ is (completely) integrable if and only if for each pair $X,Y\in\Gamma(H)$ of smooth vector fields belonging to $H$, we also have $[X,Y]\in\Gamma(H)$, i.e. $H$ is involutive.
So we would like a way to measure how much the distribution $H$ fails to be involutive. But without some extra structure we cannot assign any "tensorial" measure of involutivity.
Let this extra structure be a distribution $V\subseteq TM$ which is complementary to $H$, i.e. we have $TM=V\oplus H$. Then we have projection operators $P_V:TM\rightarrow V$ and $P_H:TM\rightarrow H$ obeying $\mathrm {id}_{TM}=P_V+P_H$.
Let's say that the complementary pair $(V,H)$ is a connection on $M$ with $H$ the horizontal distribution and $V$ the vertical distribution. In this general setting these two distributions are on completely equal footing, so the distinction of "horizontal" and "vertical" is completely arbitrary.
But now we can construct a tensor field which measures how much $H$ fails to be involutive as $$ R:\mathcal D\times\mathcal D\rightarrow\Gamma(V),\quad R(X,Y)=P_V[P_HX,P_HY], $$ where $\mathcal D=\Gamma(TM)$ is the module of smooth vector fields. This is indeed a tensor because if $f\in C^\infty(M)$ is a smooth function, we have $$ R(X,fY)=P_V[P_HX,fP_HY]=fP_V[P_HX,P_HY]+(P_HX)fP_VP_HY=fP_V[P_HX,P_HY], $$and analogously for the first slot in $R$. We can then call $R$ the curvature of the connection $(V,H)$.
Note that since $V$ is on equal footing with $H$, we can also define the cocurvature$$ Q(X,Y)=P_H[P_VX,P_VY], $$which then measures how much $V$ fails to be involutive.
We find that $H$ is involutive (and thus Frobenius-integrable) if and only if its curvature vanishes. The value of the curvature depends on the choice of $V$ but whether it is nonzero or not does not.
The situation with fibered manifolds is a special case of this one. If $\pi:N\rightarrow M$ is a fibered manifold, then the total space $N$ already has an involutive distribution $V\subseteq TN$ given by $V=\ker(T\pi)$. A connection on $(N,\pi,M)$ is then a smooth distribution $H\subseteq TN$ which is complementary to $V$. The curvature of $H$ is defined the same way it was above in the more general situation and once again $H$ is integrable if and only if the curvature vanishes. The cocurvature trivially vanishes because $V$ is always involutive.
