Not in general: here is a counterexample. Take $p = 2$, and consider the matrices $T_1 = [1,0,0; 0,0,0; 0,0,0], T_2 = [0,0,0; 0,1,0; 0,1,0], T_3 = [0,0,0; 1,0,0; 0,1,0]$. (I'm using semi-colons to separate rows; a bit of LaTeX trouble formatting the matrix ...)

Then if $v$ is the column vector $(x,y,z)$, we get that the matrix $(T_1 v, T_2 v, T_3 v)$ is $[x,0,0; 0,y,x; 0,y,y]$. It has determinant $xy^2 - yx^2$. Note that if $x,y$ are in $\mathbb{F}_2$, then $x^2 = x$ and $y^2 = y$, so this determinant is $0$. However, the polynomial $xy^2 - yx^2$ is not identically $0$, so there are values in $\overline{\mathbb{F}_2}$for which the determinant will not vanish.

Not in general: here is a counterexample. Take $p = 2$, and consider the matrices $T_1 = [1,0,0; 0,0,0; 0,0,0], T_2 = [0,0,0; 0,1,0; 0,1,0], T_3 = [0,0,0; 1,0,0; 0,1,0]$. (I'm using semi-colons to separate rows; a bit of LaTeX trouble formatting the matrix ...)

Then if $v$ is the column vector $(x,y,z)$, we get that the matrix $(T_1 v, T_2 v, T_3 v)$ is $[x,0,0; 0,y,x; 0,y,y]$. It has determinant $xy^2 - yx^2$. Note that if $x,y$ are in $\mathbb{F}_2$, then $x^2 = x$ and $y^2 = y$, so this determinant is $0$. However, the polynomial $xy^2 - yx^2$ is not identically $0$, so there are values in $\overline{\mathbb{F}_2}$for which the determinant will not vanish.

Update:

These transformations are not invertible, but you can embed this example in 4 dimensions. Namely, define $T_1(x,y,z,w) = (x,y,z,w)$, $T_2(x,y,z,w) = (x,x+y,z,w)$, $T_3(x,y,z,w) = (x,y,y+z,y+w)$ and $T_4(x,y,z,w) = (x,y,x+z, y + w)$. Then these are all clearly invertible, and the determinant of the matrix of linear forms is $x^2 y(y - x)$, which vanishes as before if $x$ and $y$ are in $\mathbb{F}_2$, but not over the algebraic closure.