What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are,
$$x_1=\sqrt{2}+\sqrt{3}$$
$$x_2=\sqrt{2}-\sqrt{3}$$
$$x_3=-\sqrt{2}+\sqrt{3}$$
$$x_4=-\sqrt{2}-\sqrt{3}$$
$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+\sqrt{3})+c(\sqrt{2}+\sqrt{3})^2+d(\sqrt{2}+\sqrt{3})^3|a,b,c,d \in \mathbb{Q}\}$. Thus an integral basis of $K$ is,
$$S=\{1,\sqrt{2}+\sqrt{3},(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3\}$$
$$S=\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},(\sqrt{2}+\sqrt{3})^3\}$$
and hence,
$$d(K)=d(\mathbb{Q}(\sqrt{2}+\sqrt{3}))=D(S)$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & (\sqrt{2}+\sqrt{3})^2 & (\sqrt{2}+\sqrt{3})^3\\
1 & \sqrt{2}-\sqrt{3} & (\sqrt{2}-\sqrt{3})^2 & (\sqrt{2}-\sqrt{3})^3\\
1 & -\sqrt{2}+\sqrt{3} & (-\sqrt{2}+\sqrt{3})^2 & (-\sqrt{2}+\sqrt{3})^3\\
1 & -\sqrt{2}-\sqrt{3} & (-\sqrt{2}-\sqrt{3})^2 & (-\sqrt{2}-\sqrt{3})^3
\end{vmatrix}^2$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & 5+2\sqrt{6} & 11\sqrt{2}+9\sqrt{3}\\
1 & \sqrt{2}-\sqrt{3} & 5-2\sqrt{6} & 11\sqrt{2}-9\sqrt{3}\\
1 & -\sqrt{2}+\sqrt{3} & 5-2\sqrt{6} & -11\sqrt{2}+9\sqrt{3}\\
1 & -\sqrt{2}-\sqrt{3} & 5+2\sqrt{6} & -11\sqrt{2}-9\sqrt{3}
\end{vmatrix}^2$$
Using Mathematica,
$$D(S)=2677248 + 1078272\sqrt{6}$$
Why am I not getting an integer answer?
Appararently I don't know how to type, $D(S)= 147456$

Well, my favorite method of defining (and calculating!) the discriminant of a free algebra is as the determinant of the trace pairing. Before I launch into this, however, in case your ring is of the form $\Bbb Q[a]$, so that a basis is the powers $\{a^m\}_{0\le m<n}$, where the rank is $n$, there’s a quick and dirty way of calculating the discriminant. This is just $\text{Norm}\bigl(f'(a)\bigr)$, where $f$ is the minimal polynomial of $a$. Using $a=\sqrt2+\sqrt3$ and its minimal polynomial, you do indeed get $2^{14}\cdot3^2$. But we know that this method is not apposite to your problem.
If it’s true that an integral basis of your field is $\{1,\sqrt2,\sqrt3, (\sqrt3-1)/\sqrt2\}$, we can label these $b_1$ through $b_4$ and write down the matrix whose $(i,j)$-th entry is $\text{Trace}(b_ib_j)$, and calculate its determinant. I’ll let you check my calculation that the matrix in question is
$$
\begin{pmatrix}
4&0&0&0\\0&8&0&-4\\0&0&12&0\\0&-4&0&8
\end{pmatrix}\,,
$$
whose determinant is $2304=2^8\cdot3^2$. Since this agrees with Daniel’s result, it seems my guess of the basis was correct.
