This is indeed a rather cumbersome calculation. Here it goes:
$$\nabla R(e_i,e_j,e_k,e_l,e_h)(p)+\nabla R(e_i,e_j,e_h,e_k,e_l)(p)+\nabla R(e_i,e_j,e_l,e_h,e_k)(p)=$$
$$\langle \nabla_{e_h} \nabla_{e_l} \nabla_{e_k}e_i - \nabla_{e_h} \nabla_{e_k} \nabla_{e_l}e_i +\nabla_{e_h} \nabla_{[e_k,e_l]}e_i$$
$$+\nabla_{e_l} \nabla_{e_k} \nabla_{e_h}e_i - \nabla_{e_l} \nabla_{e_h} \nabla_{e_k}e_i +\nabla_{e_l} \nabla_{[e_h,e_k]}e_i$$
$$+\nabla_{e_k} \nabla_{e_h} \nabla_{e_l}e_i - \nabla_{e_k} \nabla_{e_l} \nabla_{e_h}e_i +\nabla_{e_k} \nabla_{[e_l,e_h]}e_i, e_j \rangle (p) =$$
$$ \langle R(e_l,e_h)\nabla_{e_k}e_i-\nabla_{[e_l,e_h]}\nabla_{e_k}e_i+R(e_k,e_l) \nabla_{e_h}e_i -\nabla_{[e_k,e_l]}\nabla_{e_h}e_i+R(e_k,e_h)\nabla_{e_l}e_i-\nabla_{[e_k,e_h]}\nabla_{e_l}e_i+\nabla_{e_h} \nabla_{[e_k,e_l]}e_i+\nabla_{e_l} \nabla_{[e_h,e_k]}e_i+\nabla_{e_k} \nabla_{[e_l,e_h]}e_i,e_j\rangle (p)$$
$$= \langle R([e_l,e_h],e_k)e_i-\nabla_{[[e_l.e_h],e_k]}e_i+R([e_k,e_l],e_h)e_i-\nabla_{[[e_k.e_l],e_h]}e_i+R([e_k,e_h],e_l)e_i-\nabla_{[[e_k.e_h],e_l]}e_i,e_j\rangle(p) =$$
$$\langle R(\nabla_{e_l}e_h-\nabla_{e_h}e_l,e_k)e_i + R(\nabla_{e_k}e_l-\nabla_{e_l}e_k,e_h)e_i+R(\nabla_{e_k}e_h-\nabla_{e_h}e_k,e_l)e_i-
\nabla_{[[e_l.e_h],e_k]+[[e_k.e_l],e_h]+[[e_k.e_h],e_l]}e_i,e_j\rangle(p) = 0$$
In this last line the Jacoby Identity and the symmetry of the connection is used.
Observe that every time that elements of the form $\nabla_{e_l}e_h$ are dropped, what is being used is that of course $\nabla_{e_l}e_h(p)=0$, and the tensorial property of the Riemannian tensor (observe also the necessity that everything must be evaluated on p). Since p is arbitrary, and because of the linearity of everything involved, the identity is proven.