Lower bound for determinant and permanent
Permanent is complete for VNP under p-projections over any field not of characteristic 2. This provides a positive answer to your second question. If this reduction were linear, it would give a positive answer to your first question, but I believe that remains open.

In more detail: there is some polynomial $q(n)$ such that $det_n(X)$ is a projection of $perm_{q(n)}(Y)$, i.e. there is a certain substitution sending each variable $y_{ij}$ either to a variable $x_{k\ell}$ or a constant such that after this substitution the $q(n) \times q(n)$ permanent is computing the $n \times n$ determinant.

1) Thus Ryser's formula yields a depth 3 formula (depth does not increase under projections because the substitutions can be done on the input gates) of size $2^{O(q(n))}$ for determinant.   UPDATE: As @Ramprasad points out in the comments, this only gives something nontrivial if $q(n) = o(n \log n)$, since there is a trivial depth 2 formula of size $n \cdot n! = 2^{O(n \log n)}$ for det. I am with Ramprasad in that the best I know is the reduction through ABPs, which yields $q(n)=O(n^3)$. 

2) If the $m \times m$ permanent can be computed - again, over some field of characteristic not 2 - by a circuit of size $s(m)$, then $n \times n$ determinant can be computed by a circuit of size $s(q(n))$. So a lower bound of $b(n)$ on circuit-size for $det_n$ yields a lower bound of $b(q^{-1}(n))$ on circuit-size for the permanent (that's $q$ inverse, not $1/q(n)$). The above-mentioned $q(n)=O(n^3)$ yields a $b(n^{1/3})$ perm lower bound from a $b(n)$ det lower bound.