The given relations correspond to a homogeneous system of linear equations for the $9$ entries (ordered left to right, top to bottom) and the negated sum, with the following matrix:
$$
\pmatrix{
1&1&1&0&0&0&0&0&0&1\\
0&0&0&1&1&1&0&0&0&1\\
0&0&0&0&0&0&1&1&1&1\\
1&0&0&1&0&0&1&0&0&1\\
0&1&0&0&1&0&0&1&0&1\\
0&0&1&0&0&1&0&0&1&1\\
1&0&0&0&1&0&0&0&1&1\\
0&0&1&0&1&0&1&0&0&1
}
$$
This matrix has rank $7$ over the real numbers (as you can establish by hand or using Wolfram|Alpha), so the solution set is a vector space of dimension $10-7=3$. If the squares are meant to contain real numbers, their addition and scalar multiplication correspond to those of these solution vectors, so the squares also form such a vector space.
(Edit: I'm not sure if I read the OP correctly, but to my understanding, except the sums of entries along the main diagonal or main anti-diagonal, all line sums along other diagonals are not part of the definition.)
Denote by $r_i$ the $i$-th row sum, $c_j$ the $j$-th column sum, $d$ the diagonal sum and $a$ the anti-diagonal sum. Every so-called "magic square" in question must satisfy the following 9 constraints:
$$r_1=d,\ r_2=d,\ r_3=d,\ r_4=d,\ c_1=d,\ c_2=d,\ c_3=d,\ c_4=d,\ a=d.$$
(I say "so-called" in the above because the definition here deviates from the conventional one --- here, a magic square can have non-integer or even negative entries.)
Clearly the constraint $c_4=d$ is redundant, because the sum of all row sums must be equal to the sum of all column sums. Since a $4\times4$ matrix is specified by 16 entries, you now have 16 unknowns and at most 8 independent constraints. This suggests that the dimension of the vector space in question is at least 16-8=8.
So, to prove that the dimension of the vector space in question is exactly 8, you only need to show that the remaining 8 constraints are indeed linearly independent. This amounts to proving that some $8\times16$ matrix has full row rank. It takes some work but it shouldn't be hard.
Having proven that the dimension is 8, it is not hard to find a basis. All you need is to construct a magic square with nonzero row/column/diagonal/anti-diagonal sums and 7 magic squares with zero row/column/diagonal/anti-diagonal sums. This is easy:
\begin{align*}
&\pmatrix{1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1},
\ \pmatrix{1&0&0&-1\\ 0&-1&1&0\\ 0&1&-1&0\\ -1&0&0&1},\\
&\pmatrix{1&-1&0&0\\ -1&1&0&0\\ 0&0&-1&1\\ 0&0&1&-1},
\ \pmatrix{1&0&-1&0\\ 0&-1&0&1\\ -1&0&1&0\\ 0&1&0&-1},\\
&\pmatrix{0&0&-1&1\\ 0&0&1&-1\\ 1&-1&0&0\\ -1&1&0&0},
\ \pmatrix{0&-1&0&1\\ 1&0&-1&0\\ 0&1&0&-1\\ -1&0&1&0},\\
&\pmatrix{0&1&-1&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&-1&1&0},
\ \pmatrix{0&0&0&0\\ 1&0&0&-1\\ -1&0&0&1\\ 0&0&0&0}.
\end{align*}
By looking at the diagonals and anti-diagonals of their linear combinations, it should be rather obvious that these 8 magic squares are indeed linearly independent.
Best Answer
This is really an extension to my comment above. If you denote the entries of the matrix by $a,b,c,d,e,f,g,h,i$, then your condition that all rows and columns and diagonals have the same sum corresponds to the solutions to the matrix equation:
$$ \left[ \begin{array}{ccccccccc|c} 1 & 1 & 1 & 0 & 0 &0 &0 &0 &0 & k\\ 0 & 0 & 0 & 1 & 1 &1 & 0 & 0 &0 & k \\ 0 & 0 & 0 & 0& 0 &0 &1 &1 & 1 & k\\ 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0&k \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 &k\\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1&k \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 &k\\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 &k\\ \end{array} \right] $$
Where the first three rows represent the rows, the next three rows represent the columns, and the last two rows represent the diagonals. The $k$ represent any old fixed constant that you want the sums to be, all the same because this square is magic.
I'll work on reducing this matrix for this answer, but I suspect a TI84 or some other calculator can do this faster than I can.
After working this out on pen and paper and doing it improperly (of course, you can never get them right the first time), I went to this site here, and entered the matrix with $k = 1$. It doesn't matter what $k$ is since this variable is going to be free no matter what, and you can just scale that column.
The output given is this: $$ \left[ \begin{array}{ccccccccc|c} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2k/3\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 2k/3 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 &-1 &-1 & -k/3\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 &-1 &-2 & -2k/3 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & k/3\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 2 & 4k/3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & k\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{array} \right] $$
This matrix has $2$ free variables, and $k$ is also free like we said above, so this is the $3$ dimensional solution set we wanted. Since you have found $3$ such matrixes that are linearly independent and have the desired property, they must span the set of all magic squares.