We can convert `tfunc` equations to
~~~
    A0^A1^A2 = B0^(maj(A3,A4,A5)<<1)
    A1^A2^A3 = B1^(maj(B0,A6,A7)<<1)
    A2^A3^A4 = B2^(maj(B1,A0,A1)<<1)
    A3^A4^A5 = B3^(maj(B2,A2,A3)<<1)
    A4^A5^A6 = B4^(maj(B3,A4,A5)<<1)
    A5^A6^A7 = B5^(maj(B4,A6,A7)<<1)
    A6^A7^A0 = B6^(maj(B5,A0,A1)<<1)
    A7^A0^A1 = B7^(maj(B6,A2,A3)<<1)
~~~
In these, knowing the `Bj` and the low `i` bits of the `Ai` let us compute the right-hand side to `i+1` bits, thus the left-hand side to `i+1` bits, thus the `Ai` to `ì+1` bits using the invert matrix/equations already in the question.

We can apply this for `ì` from `0` to `n-1` to fully invert the function. Baring mistakes, code could be (not tested):
~~~
function inv_tfunc([B0, B1, B2, B3, B4, B5, B6, B7]) {
    M = ((1<<(n-1)-1)<<1)+1; // mask for  n  bits
    A0 = A1 = A2 = A3 = A4 = A5 = A6 = A7 = 0; // initial value is immaterial
    for (i = 0; i < n; i = i + 1) {
        C0 = (B0^(maj(A3,A4,A5)<<1))&M;
        C1 = (B1^(maj(B0,A6,A7)<<1))&M;
        C2 = (B2^(maj(B1,A0,A1)<<1))&M;
        C3 = (B3^(maj(B2,A2,A3)<<1))&M;
        C4 = (B4^(maj(B3,A4,A5)<<1))&M;
        C5 = (B5^(maj(B4,A6,A7)<<1))&M;
        C6 = (B6^(maj(B5,A0,A1)<<1))&M;
        C7 = (B7^(maj(B6,A2,A3)<<1))&M;
        A0 = C1^C2^C4^C5^C7;
        A1 = C2^C3^C5^C6^C0;
        A2 = C3^C4^C6^C7^C1;
        A3 = C4^C5^C7^C0^C2;
        A4 = C5^C6^C0^C1^C3;
        A5 = C6^C7^C1^C2^C4;
        A6 = C7^C0^C2^C3^C5;
        A7 = C0^C1^C3^C4^C6;
    }
    return [A0, A1, A2, A3, A4, A5, A6, A7];
}