Edit: I now give the argument for general reductive $G$.
Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^*(T)$ for its group of characters. Write $R$ for the
subgroup of $X$ generated by the roots of $G$. Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$.
Claim: $G$ has a faithful irreducible representation if and only if the character
group $X/R$ of $Z$ is cyclic.
Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.
In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.
As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.
For $(\Leftarrow)$ let me first treat the case where $G$ is almost simple; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.
The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi_i$
of its irreducible components. There is an isogeny
$$\pi:\prod_i G_{i,sc} \times T \to G$$
where $T$ is a torus and $G_{i,sc}$ is the simply connected almost simple group
with root system $\Phi_i$. Write $G_i$ for the image $\pi(G_{i,sc}) \subset G$.
The key fact is this:
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho_{\mid G_i}$ is non-trivial
for each $i$.
Now, as before pick $\lambda \in X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant. After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(*) \quad \text{for each $i$, there is $\beta_i \in \Phi_i$ with $\langle \lambda,\beta_i^\vee \rangle \ne 0$}$$
Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(*)$ implies that $G_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.
The center of a group and its isomorphism type can be seen from the character table: The center is the set $\newcommand{\Irr}{\operatorname{Irr}}$
$$ Z(G) = \{ g\in G \mid \: |\chi(g)| = \chi(1) \text{ for all } \chi \in \Irr(G)\} .$$
Since $\chi_{Z(G)}=\chi(1)\lambda$, you see the linear characters of $Z(G)$ in the character table, and thus you see the isomorphism type of $Z(G)$.
The irreps lying over two different characters of $Z(G)$ need not be related. To see this, consider a direct product $G=H\times K$. Then $Z(G)=Z(H)\times Z(K)$. The irreps of $G$ lying over $\lambda \times \mu$ are tensors of irreps of $H$ over $\lambda$ with irreps of $K$ over $\mu$. For characters, this reads
$$\Irr(G\mid \lambda\times \mu) = \Irr(H\mid\lambda)\times \Irr(K\mid\mu).$$
Now taking $\lambda\neq 1 = \mu$ or vice versa and suitable examples for $H$ and $K$, we see that these sets can look quite different.
An interesting fact is that the linear characters of $Z(G)$ yield a grading of $\Irr(G)$: the linear characters define a partition of $\Irr(G)$ and if $\chi\in \Irr(G)$ lies over $\lambda$ and $\psi$ over $\mu$, then all irred. constituents of $\chi\phi$ lie over $\lambda \mu$. Moreover, it is not too difficult to show that every grading of $\Irr(G)$ by some group comes in fact from a subgroup $Z$ of the center, that is, $\chi$ and $\psi$ are in the same subset of the partition if the restrictions $\chi_Z$ and $\psi_Z$ contain the same character of $Z$. (This idea has been used by Gelaki and Nikshych to define nilpotency of arbitrary fusion categories: arXiv:math/0610726)
Best Answer
The irreducible character degrees for this group have degree $1,3,3,6,7,8$.
Later edit: Note that for the irreducible characters of degree $6,7$ and $8$ the representations above may be explicitly given as real representations. The degree $8$ representation requires a little further thought. If we induce the trivial character from a Sylow $2$-subgroup, the $8$-dimensional irreducible character occurs with multiplicity $1$, the trivial character occurs once, the $7$-dimensional irreducible character does not occur, the $6$-dimensional character occurs once and the two three dimensional character each occur once. Since the permutation representation is realized over $\mathbb{R},$ it follows that the $8$-dimensional representation may be realized over the real field.
The two $3$-dimensional representations do not have real characters. To obtain a real representation of their sum, do the last procedure instead with a real irreducible two dimensional orthogonal representation of the Sylow $7$-subgroup (which is just as group of real rotations): two copies of each of the $6,7$ and $8$ degree real orthogonal representations show up, so take the orthogonal complement of their sum, obtaining an orthogonal representation of degree $6$ which is irreducible as a real representation, but not a complex representation.