In Massey's paper "On the Stiefel Whitney classes of a manifold I" he shows that manifolds of dimension n = 4s + 3 have $w_n = w_{n-1} = w_{n-2} = 0$. Where $w_i$ is a mod 2 Stiefel-Whitney class
Also, the first non-zero Stiefel-Whitney class must be $w_{2^k}$ for some k. That means that if $M$ is a 7-dimensinal Spin manifold, the only class which has any hope of not being zero is $w_4$.
In general, it doesn't seem like $w_4$ has any "special" meaning (like $w_2$ tells about spin and $w_1$ orientability) but maybe in this specific case it does. Is there a condition that tells us when $w_4$ must be nonzero for 7-d Spin manifolds?
Here is an idea: Suppose that $w_4 \neq 0$. The Poincare duality tells us that there is a nonzero class in $H^3(M)$. Do we know anything about that three class? Maybe it has to be zero, implying $w_4 = 0$?
You know that $\text{Sq}(\nu) = w$, where $\nu$ is the Wu class. You also know that because $M$ is a spin manifold, you have $w_1(M) = w_2(M) = 0$, and thus as you say $w_3(M) = 0$. This implies that $\nu_1 = \nu_2 = \nu_3 = 0$ as well.
Now recall that $\nu_k$ has the property that $\langle \nu_k \smile c, [M]\rangle = \langle \text{Sq}^k c, [M]\rangle$ for any class $c \in H^{\dim M - k}(M;\Bbb Z/2)$. If $k > \dim M/2$, then $k > \dim M - k$, so that $\text{Sq}^k c = 0$ automatically (this is a defining property of the Steenrod square). Thus $\nu_k = 0$ for all $k > \dim M/2$.
For spin 7-manifolds, this means that the total Wu class is $\nu = 1$ and hence $w = \text{Sq}(1) = 1$ --- so all Stiefel-Whitney classes are automatically zero as well.