You used the tables for I/O loads correctly, but you forgot to apply the frequency scaling, table shown below:
N | fs
---------
2 | 1.36
3 | 1.75
4 | 2.13
5 | 2.42
6 | 2.7
7 | 2.95
8 | 3.17
9 | 3.39
10 | 3.58
If you divide your values to 1.36 you'll see that the results match.
Passive filter synthesis is a large domain but, in general, the filter is treated like a two-port network, having a voltage source with a series resistor as an input, and a resistor at the output (doubly-terminated passive port). Part of the design procedure is determining what elements the port will have based on a transfer function.
For this case, the requirements are \$R_L=1, R_S=\frac12\$, and a 2nd order Bessel with a group delay of \$\tau=8\text{ms}\$. That makes the whole circuit look like this:
I'll take a different approach in finding the elements. The transfer functions for a 2nd order, normalized, prototype lowpass Bessel, and the circuit above are:
$$\begin{align}
H(s)&=\frac{3}{s^2+3s+3}=\frac{1}{\frac{s^2}{3}+s+1} \\
G(s)&=\frac{R_L}{R_LLCs^2+(L+R_SR_LC)s+R_S+R_L}=\frac{1}{LCs^2+(L+0.5C)s+1.5}
\end{align}$$
It's apparent that \$\small G(0)=\frac23\$, so \$\small G(s)=\frac23H(s)\$. I've chosen the non-monic representation because now it's easy to make a system of equations from the denominators, only, from which the values of the elements can be readily found, for two values of \$s\$ (which don't need to be complex). Since for \$s=0\$ both \$L\$ and \$C\$ disappear, let's choose \$s=1\$ and \$s=2\$:
$$\begin{align}\left\{
\begin{array}{x}
G(1)=\frac23H(1)\quad\Rightarrow\quad LC+L+C+1.5&=1.5(\frac13+1+1) \\
G(2)=\frac23H(2)\quad\Rightarrow\quad 4LC+2(L+C)+1.5&=1.5(\frac43+2+1)
\end{array}\right.
\end{align} \\
\Rightarrow \\
\begin{array}{x}
L=1.31\,&,\quad 0.191 \\
C=0.382\,&,\quad 2.618
\end{array}$$
The solution shows two sets of values, any of them can be chosen (shown below). As mentioned in the comments, the frequency scaling which you had to apply was simply the frequency of the lowpass prototype when:
$$
|H(j\omega)|=\left|\frac{3}{-\omega^2+j3\omega+3}\right|=\frac{3}{\sqrt{\left(3-\omega^2\right)^2+9\omega^2}}=\frac{1}{\sqrt{2}} \\
\Rightarrow \\
\omega_1=\pm\sqrt{\frac32}\sqrt{\sqrt5-1}\quad\omega_2=\pm j\sqrt{\frac32}\sqrt{\sqrt5+1}$$
Since we're dealing with positive, real numbers, \$\omega\approx 1.36\$ remains, which is what is given in the table. The same goes for all the orders.
Now, if you take the value of the normalized elements from the table, \$L=\frac{0.2601}{1.36}=0.19125\$ and \$C=\frac{3.5649}{1.36}=2.62125\$, values which are very close to the ones calculated above, save roundings & co. This can be applied to higher orders; the system of equations will get fluffy, but it can be done. A quick check shows that the results are valid:
V(a) is the magnitude-scaled lowpass prototype, V(b) is your corrected result, V(c) and V(d) are the two solutions. The traces have been slightly displaced to avoid overlapping, but they are the same, as shown by the group delay (dotted trace). Whatever minor differences exist, they're due to rounding, nothing more. If they had been calculated with float precision, or more, they would have overlapped perfectly.
For the sake of proof, here's how a 3rd order Bessel could be deduced in the same way (\$R_S=2, R_L=3, \tau=1\text{s}\$):
$$\begin{align}
H(s)&=\frac{15}{s^3+6*s^2+15s+15} \\
G(s)&=\frac{R_L}{L_1L_2Cs^3+(R_SL_2C+R_LL_1C)s^2+(R_LR_SC+L_1+L_2)s+R_S+R_L}
\end{align}$$
The system of equations can be built in a similar manner and there will be a lot of solutions (32), but many of them will have negative or complex values. Sorting them out leaves these two:
$$\left\{
\begin{array}{x}
L_1=0.404\,&,\quad 3.378 \\
L_2=2.87\,&,\quad 0.555 \\
C=0.288\,&,\quad 0.178
\end{array}\right.$$
Both solutions work (this time I let all traces overlap):
