Of course we can!
And if we actually do, then we find that $\phi(t, x) = |\nabla_x p_t^{(n)}(x)|$ satisfies $$\phi(t, x) = t^{-(n+1)/(2s)} \phi(1, t^{-1/(2s)} x)$$ and, with $\tilde x = (x, 0, 0) \in \mathbb R^{n+2}$), $$\phi(1,x) = |x| p_1^{(n+2)}(\tilde x) \approx C |x| \min\{1, |x|^{-(n+2+2s)}\} = C \min\{|x|, |x|^{-(n+1+2s)}\} $$ (here we use the well-known fact that $p_1^{(n)}(x) \approx C \min\{1, |x|^{-n-2s}\}$). It follows that $$\|\phi(1,\cdot)\|_q < \infty \qquad \text{if and only if } q > \tfrac{n}{n+1+2s} $$ and $$\|\phi(t,\cdot)\|_q = t^{-(n+1)/2s + n/(2qs)} ,$$ and the above is in $L^p([0,T])$ if and only if $$ p (\tfrac{n+1}{2s} - \tfrac{n}{2qs}) < 1 . $$ So the answer seems to be rather ugly: $$ q > \tfrac{n}{n+1+2s} \qquad \text{and} \qquad p (\tfrac{n+1}{2s} - \tfrac{n}{2qs}) < 1 , $$ unless I made some mistake.